Value of $lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+…+sqrt{n-1}}{nsqrt{n}}$












3












$begingroup$



Evaluate
$$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}}.$$




I am trying to use the Sandwich principle here..



$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}}=lim_{n to infty}dfrac{sqrt{dfrac{1}{n}}+sqrt{dfrac{2}{n}}+sqrt{dfrac{3}{n}}+...+sqrt{dfrac{n-1}{n}}}{n}ge lim_{n to infty}bigg(sqrt{dfrac{1}{n}}.sqrt{dfrac{2}{n}}.sqrt{dfrac{3}{n}}....sqrt{dfrac{n-1}{n}}bigg )^dfrac{1}{n-1}=lim_{n to infty}bigg(dfrac{(n-1)!}{n^n}bigg )^dfrac{1}{2(n-1)}$



But after this I am a little in doubt.
This link may provide some light Evaluation of the limit $limlimits_{n to infty } frac1{sqrt n}left(1 + frac1{sqrt 2 }+frac1{sqrt 3 }+cdots+frac1{sqrt n } right)$ but I do not understand how it would help my problem..



In continuation of @Rebello's answer here, I would like to provide an answer for the problem given in the link



$limlimits_{n to infty } frac1{sqrt n}left(1 + frac1{sqrt 2 }+frac1{sqrt 3 }+cdots+frac1{sqrt n } right)=limlimits_{n to infty }sum_{k=1}^{n}{dfrac{1}{sqrt{kn}}}=int_{0}^{1}dfrac{1}{sqrt{x}}dx+limlimits_{n to infty }dfrac{1}{n}=2$










share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Well, this is just a Riemann sum.
    $endgroup$
    – Zacky
    Dec 15 '18 at 8:45






  • 1




    $begingroup$
    Oh is it? Leave it then I can do this... I forgot Reimann's sum. Sorry
    $endgroup$
    – Saradamani
    Dec 15 '18 at 8:46






  • 2




    $begingroup$
    Okay then :D Also post it as answer when you finish solving it.
    $endgroup$
    – Zacky
    Dec 15 '18 at 8:48












  • $begingroup$
    There are $n-1$ terms in the numerator
    $endgroup$
    – Shubham Johri
    Dec 15 '18 at 8:51
















3












$begingroup$



Evaluate
$$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}}.$$




I am trying to use the Sandwich principle here..



$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}}=lim_{n to infty}dfrac{sqrt{dfrac{1}{n}}+sqrt{dfrac{2}{n}}+sqrt{dfrac{3}{n}}+...+sqrt{dfrac{n-1}{n}}}{n}ge lim_{n to infty}bigg(sqrt{dfrac{1}{n}}.sqrt{dfrac{2}{n}}.sqrt{dfrac{3}{n}}....sqrt{dfrac{n-1}{n}}bigg )^dfrac{1}{n-1}=lim_{n to infty}bigg(dfrac{(n-1)!}{n^n}bigg )^dfrac{1}{2(n-1)}$



But after this I am a little in doubt.
This link may provide some light Evaluation of the limit $limlimits_{n to infty } frac1{sqrt n}left(1 + frac1{sqrt 2 }+frac1{sqrt 3 }+cdots+frac1{sqrt n } right)$ but I do not understand how it would help my problem..



In continuation of @Rebello's answer here, I would like to provide an answer for the problem given in the link



$limlimits_{n to infty } frac1{sqrt n}left(1 + frac1{sqrt 2 }+frac1{sqrt 3 }+cdots+frac1{sqrt n } right)=limlimits_{n to infty }sum_{k=1}^{n}{dfrac{1}{sqrt{kn}}}=int_{0}^{1}dfrac{1}{sqrt{x}}dx+limlimits_{n to infty }dfrac{1}{n}=2$










share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Well, this is just a Riemann sum.
    $endgroup$
    – Zacky
    Dec 15 '18 at 8:45






  • 1




    $begingroup$
    Oh is it? Leave it then I can do this... I forgot Reimann's sum. Sorry
    $endgroup$
    – Saradamani
    Dec 15 '18 at 8:46






  • 2




    $begingroup$
    Okay then :D Also post it as answer when you finish solving it.
    $endgroup$
    – Zacky
    Dec 15 '18 at 8:48












  • $begingroup$
    There are $n-1$ terms in the numerator
    $endgroup$
    – Shubham Johri
    Dec 15 '18 at 8:51














3












3








3


4



$begingroup$



Evaluate
$$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}}.$$




I am trying to use the Sandwich principle here..



$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}}=lim_{n to infty}dfrac{sqrt{dfrac{1}{n}}+sqrt{dfrac{2}{n}}+sqrt{dfrac{3}{n}}+...+sqrt{dfrac{n-1}{n}}}{n}ge lim_{n to infty}bigg(sqrt{dfrac{1}{n}}.sqrt{dfrac{2}{n}}.sqrt{dfrac{3}{n}}....sqrt{dfrac{n-1}{n}}bigg )^dfrac{1}{n-1}=lim_{n to infty}bigg(dfrac{(n-1)!}{n^n}bigg )^dfrac{1}{2(n-1)}$



But after this I am a little in doubt.
This link may provide some light Evaluation of the limit $limlimits_{n to infty } frac1{sqrt n}left(1 + frac1{sqrt 2 }+frac1{sqrt 3 }+cdots+frac1{sqrt n } right)$ but I do not understand how it would help my problem..



In continuation of @Rebello's answer here, I would like to provide an answer for the problem given in the link



$limlimits_{n to infty } frac1{sqrt n}left(1 + frac1{sqrt 2 }+frac1{sqrt 3 }+cdots+frac1{sqrt n } right)=limlimits_{n to infty }sum_{k=1}^{n}{dfrac{1}{sqrt{kn}}}=int_{0}^{1}dfrac{1}{sqrt{x}}dx+limlimits_{n to infty }dfrac{1}{n}=2$










share|cite|improve this question











$endgroup$





Evaluate
$$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}}.$$




I am trying to use the Sandwich principle here..



$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}}=lim_{n to infty}dfrac{sqrt{dfrac{1}{n}}+sqrt{dfrac{2}{n}}+sqrt{dfrac{3}{n}}+...+sqrt{dfrac{n-1}{n}}}{n}ge lim_{n to infty}bigg(sqrt{dfrac{1}{n}}.sqrt{dfrac{2}{n}}.sqrt{dfrac{3}{n}}....sqrt{dfrac{n-1}{n}}bigg )^dfrac{1}{n-1}=lim_{n to infty}bigg(dfrac{(n-1)!}{n^n}bigg )^dfrac{1}{2(n-1)}$



But after this I am a little in doubt.
This link may provide some light Evaluation of the limit $limlimits_{n to infty } frac1{sqrt n}left(1 + frac1{sqrt 2 }+frac1{sqrt 3 }+cdots+frac1{sqrt n } right)$ but I do not understand how it would help my problem..



In continuation of @Rebello's answer here, I would like to provide an answer for the problem given in the link



$limlimits_{n to infty } frac1{sqrt n}left(1 + frac1{sqrt 2 }+frac1{sqrt 3 }+cdots+frac1{sqrt n } right)=limlimits_{n to infty }sum_{k=1}^{n}{dfrac{1}{sqrt{kn}}}=int_{0}^{1}dfrac{1}{sqrt{x}}dx+limlimits_{n to infty }dfrac{1}{n}=2$







calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 9:46









Robert Z

99.8k1068140




99.8k1068140










asked Dec 15 '18 at 8:42









SaradamaniSaradamani

770314




770314








  • 4




    $begingroup$
    Well, this is just a Riemann sum.
    $endgroup$
    – Zacky
    Dec 15 '18 at 8:45






  • 1




    $begingroup$
    Oh is it? Leave it then I can do this... I forgot Reimann's sum. Sorry
    $endgroup$
    – Saradamani
    Dec 15 '18 at 8:46






  • 2




    $begingroup$
    Okay then :D Also post it as answer when you finish solving it.
    $endgroup$
    – Zacky
    Dec 15 '18 at 8:48












  • $begingroup$
    There are $n-1$ terms in the numerator
    $endgroup$
    – Shubham Johri
    Dec 15 '18 at 8:51














  • 4




    $begingroup$
    Well, this is just a Riemann sum.
    $endgroup$
    – Zacky
    Dec 15 '18 at 8:45






  • 1




    $begingroup$
    Oh is it? Leave it then I can do this... I forgot Reimann's sum. Sorry
    $endgroup$
    – Saradamani
    Dec 15 '18 at 8:46






  • 2




    $begingroup$
    Okay then :D Also post it as answer when you finish solving it.
    $endgroup$
    – Zacky
    Dec 15 '18 at 8:48












  • $begingroup$
    There are $n-1$ terms in the numerator
    $endgroup$
    – Shubham Johri
    Dec 15 '18 at 8:51








4




4




$begingroup$
Well, this is just a Riemann sum.
$endgroup$
– Zacky
Dec 15 '18 at 8:45




$begingroup$
Well, this is just a Riemann sum.
$endgroup$
– Zacky
Dec 15 '18 at 8:45




1




1




$begingroup$
Oh is it? Leave it then I can do this... I forgot Reimann's sum. Sorry
$endgroup$
– Saradamani
Dec 15 '18 at 8:46




$begingroup$
Oh is it? Leave it then I can do this... I forgot Reimann's sum. Sorry
$endgroup$
– Saradamani
Dec 15 '18 at 8:46




2




2




$begingroup$
Okay then :D Also post it as answer when you finish solving it.
$endgroup$
– Zacky
Dec 15 '18 at 8:48






$begingroup$
Okay then :D Also post it as answer when you finish solving it.
$endgroup$
– Zacky
Dec 15 '18 at 8:48














$begingroup$
There are $n-1$ terms in the numerator
$endgroup$
– Shubham Johri
Dec 15 '18 at 8:51




$begingroup$
There are $n-1$ terms in the numerator
$endgroup$
– Shubham Johri
Dec 15 '18 at 8:51










4 Answers
4






active

oldest

votes


















13












$begingroup$

Hint :



$$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}} = lim_{n to infty} frac{1}{n} sum_{k=0}^{n-1} sqrt{frac{k}{n}} = int_0^1sqrt{x}mathrm{d}x$$






share|cite|improve this answer









$endgroup$





















    6












    $begingroup$

    Just for your curiosity since you already received good answers.



    We can approximate the value of
    $$a_n=frac{sum_{i=1}^{n-1} sqrt i }{n sqrt n}$$
    $$sum_{i=1}^{n-1} sqrt i =sum_{i=1}^{n} sqrt i -sqrt n=H_n^{left(-frac{1}{2}right)}-sqrt n$$ where appear generalized harmonic numbers.



    Now, using the asymptotics
    $$H_n^{left(-frac{1}{2}right)}=frac{2 nsqrt n}{3}+frac{sqrt{n}}{2}+zeta
    left(-frac{1}{2}right)+frac{1}{24sqrt n}+Oleft(frac{1}{n^{5/2}}
    right)$$
    making, for large $n$
    $$a_n=frac{frac{2 nsqrt n}{3}-frac{sqrt{n}}{2}+zeta
    left(-frac{1}{2}right)+frac{1}{24sqrt n}+Oleft(frac{1}{n^{5/2}}
    right)} {n sqrt n}=frac{2}{3}-frac{1}{2 n}+frac{zeta
    left(-frac{1}{2}right) } {n sqrt n }+Oleft(frac{1}{n^{2}}
    right)$$
    which, for sure, shows the limit and also how it is approached.



    But it also gives a good approximation even for small values of $n$. Using $zeta
    left(-frac{1}{2}right)approx -0.207886$
    and $n=10$, this would give $a_{10}approx 0.610093$ while the "exact" value is $a_{10}approx 0.610509$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      It is a fantastically shown method to tackle the problem. I heard of this approximation of Harmonic series in discrete mathematics. But never heard of $(H_{n})^{-dfrac{1}{2}}$. That's something I have known only today.
      $endgroup$
      – Saradamani
      Dec 16 '18 at 5:56





















    5












    $begingroup$

    One more approach. By Stolz-Cesaro Theorem
    $$begin{align}lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}}{nsqrt{n}} &= lim_{n to infty} frac{sqrt{n}}{(n+1)sqrt{n+1}-nsqrt{n}}
    \&=lim_{n to infty} frac{1}{(n+1)sqrt{1+frac{1}{n}}-n}\
    &=frac{1}{1+frac{1}{2}}=frac{2}{3}end{align}$$

    where we used the fact that $sqrt{1+frac{1}{n}}=1+frac{1}{2n}+o(1/n)$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But I do not understand how this Stolz-Cesaro Theorem would help me get the $dfrac{sqrt{n}}{(n+1){sqrt{n+1}}-nsqrt{n}}$ .Please elaborate because according to this theorem if $lim_{n to infty}dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}=l implies lim_{n to infty} dfrac{a_n}{b_n}=l $.But how does this relate to the fraction you obtained. Please dont mind my stupidity for not understanding this..
      $endgroup$
      – Saradamani
      Dec 15 '18 at 16:35












    • $begingroup$
      @Saradamani I added a few details. Is it clear now?
      $endgroup$
      – Robert Z
      Dec 15 '18 at 17:29






    • 1




      $begingroup$
      @Saradamani For Stolz-Cesaro we have $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ and $b_n=nsqrt{n}$.
      $endgroup$
      – Robert Z
      Dec 15 '18 at 17:31






    • 1




      $begingroup$
      No, If $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ then $a_{n+1}-a_n=(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}+sqrt{n})-(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1})=sqrt{n}$
      $endgroup$
      – Robert Z
      Dec 16 '18 at 6:38






    • 1




      $begingroup$
      @Saradamani As a matter of fact, the classic proof of Stolz-Cesaro Theorem (see link above) is based on the sandwich principle...
      $endgroup$
      – Robert Z
      Dec 16 '18 at 7:05





















    0












    $begingroup$

    It is sufficient to obtain integral of $sqrt{x}$, when $0leq xleq 1$.






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3040278%2fvalue-of-lim-n-to-infty-dfrac-sqrt1-sqrt2-sqrt3-sqrtn-1n%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      13












      $begingroup$

      Hint :



      $$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}} = lim_{n to infty} frac{1}{n} sum_{k=0}^{n-1} sqrt{frac{k}{n}} = int_0^1sqrt{x}mathrm{d}x$$






      share|cite|improve this answer









      $endgroup$


















        13












        $begingroup$

        Hint :



        $$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}} = lim_{n to infty} frac{1}{n} sum_{k=0}^{n-1} sqrt{frac{k}{n}} = int_0^1sqrt{x}mathrm{d}x$$






        share|cite|improve this answer









        $endgroup$
















          13












          13








          13





          $begingroup$

          Hint :



          $$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}} = lim_{n to infty} frac{1}{n} sum_{k=0}^{n-1} sqrt{frac{k}{n}} = int_0^1sqrt{x}mathrm{d}x$$






          share|cite|improve this answer









          $endgroup$



          Hint :



          $$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}} = lim_{n to infty} frac{1}{n} sum_{k=0}^{n-1} sqrt{frac{k}{n}} = int_0^1sqrt{x}mathrm{d}x$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 8:51









          RebellosRebellos

          15.1k31250




          15.1k31250























              6












              $begingroup$

              Just for your curiosity since you already received good answers.



              We can approximate the value of
              $$a_n=frac{sum_{i=1}^{n-1} sqrt i }{n sqrt n}$$
              $$sum_{i=1}^{n-1} sqrt i =sum_{i=1}^{n} sqrt i -sqrt n=H_n^{left(-frac{1}{2}right)}-sqrt n$$ where appear generalized harmonic numbers.



              Now, using the asymptotics
              $$H_n^{left(-frac{1}{2}right)}=frac{2 nsqrt n}{3}+frac{sqrt{n}}{2}+zeta
              left(-frac{1}{2}right)+frac{1}{24sqrt n}+Oleft(frac{1}{n^{5/2}}
              right)$$
              making, for large $n$
              $$a_n=frac{frac{2 nsqrt n}{3}-frac{sqrt{n}}{2}+zeta
              left(-frac{1}{2}right)+frac{1}{24sqrt n}+Oleft(frac{1}{n^{5/2}}
              right)} {n sqrt n}=frac{2}{3}-frac{1}{2 n}+frac{zeta
              left(-frac{1}{2}right) } {n sqrt n }+Oleft(frac{1}{n^{2}}
              right)$$
              which, for sure, shows the limit and also how it is approached.



              But it also gives a good approximation even for small values of $n$. Using $zeta
              left(-frac{1}{2}right)approx -0.207886$
              and $n=10$, this would give $a_{10}approx 0.610093$ while the "exact" value is $a_{10}approx 0.610509$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                It is a fantastically shown method to tackle the problem. I heard of this approximation of Harmonic series in discrete mathematics. But never heard of $(H_{n})^{-dfrac{1}{2}}$. That's something I have known only today.
                $endgroup$
                – Saradamani
                Dec 16 '18 at 5:56


















              6












              $begingroup$

              Just for your curiosity since you already received good answers.



              We can approximate the value of
              $$a_n=frac{sum_{i=1}^{n-1} sqrt i }{n sqrt n}$$
              $$sum_{i=1}^{n-1} sqrt i =sum_{i=1}^{n} sqrt i -sqrt n=H_n^{left(-frac{1}{2}right)}-sqrt n$$ where appear generalized harmonic numbers.



              Now, using the asymptotics
              $$H_n^{left(-frac{1}{2}right)}=frac{2 nsqrt n}{3}+frac{sqrt{n}}{2}+zeta
              left(-frac{1}{2}right)+frac{1}{24sqrt n}+Oleft(frac{1}{n^{5/2}}
              right)$$
              making, for large $n$
              $$a_n=frac{frac{2 nsqrt n}{3}-frac{sqrt{n}}{2}+zeta
              left(-frac{1}{2}right)+frac{1}{24sqrt n}+Oleft(frac{1}{n^{5/2}}
              right)} {n sqrt n}=frac{2}{3}-frac{1}{2 n}+frac{zeta
              left(-frac{1}{2}right) } {n sqrt n }+Oleft(frac{1}{n^{2}}
              right)$$
              which, for sure, shows the limit and also how it is approached.



              But it also gives a good approximation even for small values of $n$. Using $zeta
              left(-frac{1}{2}right)approx -0.207886$
              and $n=10$, this would give $a_{10}approx 0.610093$ while the "exact" value is $a_{10}approx 0.610509$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                It is a fantastically shown method to tackle the problem. I heard of this approximation of Harmonic series in discrete mathematics. But never heard of $(H_{n})^{-dfrac{1}{2}}$. That's something I have known only today.
                $endgroup$
                – Saradamani
                Dec 16 '18 at 5:56
















              6












              6








              6





              $begingroup$

              Just for your curiosity since you already received good answers.



              We can approximate the value of
              $$a_n=frac{sum_{i=1}^{n-1} sqrt i }{n sqrt n}$$
              $$sum_{i=1}^{n-1} sqrt i =sum_{i=1}^{n} sqrt i -sqrt n=H_n^{left(-frac{1}{2}right)}-sqrt n$$ where appear generalized harmonic numbers.



              Now, using the asymptotics
              $$H_n^{left(-frac{1}{2}right)}=frac{2 nsqrt n}{3}+frac{sqrt{n}}{2}+zeta
              left(-frac{1}{2}right)+frac{1}{24sqrt n}+Oleft(frac{1}{n^{5/2}}
              right)$$
              making, for large $n$
              $$a_n=frac{frac{2 nsqrt n}{3}-frac{sqrt{n}}{2}+zeta
              left(-frac{1}{2}right)+frac{1}{24sqrt n}+Oleft(frac{1}{n^{5/2}}
              right)} {n sqrt n}=frac{2}{3}-frac{1}{2 n}+frac{zeta
              left(-frac{1}{2}right) } {n sqrt n }+Oleft(frac{1}{n^{2}}
              right)$$
              which, for sure, shows the limit and also how it is approached.



              But it also gives a good approximation even for small values of $n$. Using $zeta
              left(-frac{1}{2}right)approx -0.207886$
              and $n=10$, this would give $a_{10}approx 0.610093$ while the "exact" value is $a_{10}approx 0.610509$.






              share|cite|improve this answer









              $endgroup$



              Just for your curiosity since you already received good answers.



              We can approximate the value of
              $$a_n=frac{sum_{i=1}^{n-1} sqrt i }{n sqrt n}$$
              $$sum_{i=1}^{n-1} sqrt i =sum_{i=1}^{n} sqrt i -sqrt n=H_n^{left(-frac{1}{2}right)}-sqrt n$$ where appear generalized harmonic numbers.



              Now, using the asymptotics
              $$H_n^{left(-frac{1}{2}right)}=frac{2 nsqrt n}{3}+frac{sqrt{n}}{2}+zeta
              left(-frac{1}{2}right)+frac{1}{24sqrt n}+Oleft(frac{1}{n^{5/2}}
              right)$$
              making, for large $n$
              $$a_n=frac{frac{2 nsqrt n}{3}-frac{sqrt{n}}{2}+zeta
              left(-frac{1}{2}right)+frac{1}{24sqrt n}+Oleft(frac{1}{n^{5/2}}
              right)} {n sqrt n}=frac{2}{3}-frac{1}{2 n}+frac{zeta
              left(-frac{1}{2}right) } {n sqrt n }+Oleft(frac{1}{n^{2}}
              right)$$
              which, for sure, shows the limit and also how it is approached.



              But it also gives a good approximation even for small values of $n$. Using $zeta
              left(-frac{1}{2}right)approx -0.207886$
              and $n=10$, this would give $a_{10}approx 0.610093$ while the "exact" value is $a_{10}approx 0.610509$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 15 '18 at 9:38









              Claude LeiboviciClaude Leibovici

              123k1157135




              123k1157135












              • $begingroup$
                It is a fantastically shown method to tackle the problem. I heard of this approximation of Harmonic series in discrete mathematics. But never heard of $(H_{n})^{-dfrac{1}{2}}$. That's something I have known only today.
                $endgroup$
                – Saradamani
                Dec 16 '18 at 5:56




















              • $begingroup$
                It is a fantastically shown method to tackle the problem. I heard of this approximation of Harmonic series in discrete mathematics. But never heard of $(H_{n})^{-dfrac{1}{2}}$. That's something I have known only today.
                $endgroup$
                – Saradamani
                Dec 16 '18 at 5:56


















              $begingroup$
              It is a fantastically shown method to tackle the problem. I heard of this approximation of Harmonic series in discrete mathematics. But never heard of $(H_{n})^{-dfrac{1}{2}}$. That's something I have known only today.
              $endgroup$
              – Saradamani
              Dec 16 '18 at 5:56






              $begingroup$
              It is a fantastically shown method to tackle the problem. I heard of this approximation of Harmonic series in discrete mathematics. But never heard of $(H_{n})^{-dfrac{1}{2}}$. That's something I have known only today.
              $endgroup$
              – Saradamani
              Dec 16 '18 at 5:56













              5












              $begingroup$

              One more approach. By Stolz-Cesaro Theorem
              $$begin{align}lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}}{nsqrt{n}} &= lim_{n to infty} frac{sqrt{n}}{(n+1)sqrt{n+1}-nsqrt{n}}
              \&=lim_{n to infty} frac{1}{(n+1)sqrt{1+frac{1}{n}}-n}\
              &=frac{1}{1+frac{1}{2}}=frac{2}{3}end{align}$$

              where we used the fact that $sqrt{1+frac{1}{n}}=1+frac{1}{2n}+o(1/n)$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                But I do not understand how this Stolz-Cesaro Theorem would help me get the $dfrac{sqrt{n}}{(n+1){sqrt{n+1}}-nsqrt{n}}$ .Please elaborate because according to this theorem if $lim_{n to infty}dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}=l implies lim_{n to infty} dfrac{a_n}{b_n}=l $.But how does this relate to the fraction you obtained. Please dont mind my stupidity for not understanding this..
                $endgroup$
                – Saradamani
                Dec 15 '18 at 16:35












              • $begingroup$
                @Saradamani I added a few details. Is it clear now?
                $endgroup$
                – Robert Z
                Dec 15 '18 at 17:29






              • 1




                $begingroup$
                @Saradamani For Stolz-Cesaro we have $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ and $b_n=nsqrt{n}$.
                $endgroup$
                – Robert Z
                Dec 15 '18 at 17:31






              • 1




                $begingroup$
                No, If $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ then $a_{n+1}-a_n=(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}+sqrt{n})-(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1})=sqrt{n}$
                $endgroup$
                – Robert Z
                Dec 16 '18 at 6:38






              • 1




                $begingroup$
                @Saradamani As a matter of fact, the classic proof of Stolz-Cesaro Theorem (see link above) is based on the sandwich principle...
                $endgroup$
                – Robert Z
                Dec 16 '18 at 7:05


















              5












              $begingroup$

              One more approach. By Stolz-Cesaro Theorem
              $$begin{align}lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}}{nsqrt{n}} &= lim_{n to infty} frac{sqrt{n}}{(n+1)sqrt{n+1}-nsqrt{n}}
              \&=lim_{n to infty} frac{1}{(n+1)sqrt{1+frac{1}{n}}-n}\
              &=frac{1}{1+frac{1}{2}}=frac{2}{3}end{align}$$

              where we used the fact that $sqrt{1+frac{1}{n}}=1+frac{1}{2n}+o(1/n)$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                But I do not understand how this Stolz-Cesaro Theorem would help me get the $dfrac{sqrt{n}}{(n+1){sqrt{n+1}}-nsqrt{n}}$ .Please elaborate because according to this theorem if $lim_{n to infty}dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}=l implies lim_{n to infty} dfrac{a_n}{b_n}=l $.But how does this relate to the fraction you obtained. Please dont mind my stupidity for not understanding this..
                $endgroup$
                – Saradamani
                Dec 15 '18 at 16:35












              • $begingroup$
                @Saradamani I added a few details. Is it clear now?
                $endgroup$
                – Robert Z
                Dec 15 '18 at 17:29






              • 1




                $begingroup$
                @Saradamani For Stolz-Cesaro we have $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ and $b_n=nsqrt{n}$.
                $endgroup$
                – Robert Z
                Dec 15 '18 at 17:31






              • 1




                $begingroup$
                No, If $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ then $a_{n+1}-a_n=(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}+sqrt{n})-(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1})=sqrt{n}$
                $endgroup$
                – Robert Z
                Dec 16 '18 at 6:38






              • 1




                $begingroup$
                @Saradamani As a matter of fact, the classic proof of Stolz-Cesaro Theorem (see link above) is based on the sandwich principle...
                $endgroup$
                – Robert Z
                Dec 16 '18 at 7:05
















              5












              5








              5





              $begingroup$

              One more approach. By Stolz-Cesaro Theorem
              $$begin{align}lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}}{nsqrt{n}} &= lim_{n to infty} frac{sqrt{n}}{(n+1)sqrt{n+1}-nsqrt{n}}
              \&=lim_{n to infty} frac{1}{(n+1)sqrt{1+frac{1}{n}}-n}\
              &=frac{1}{1+frac{1}{2}}=frac{2}{3}end{align}$$

              where we used the fact that $sqrt{1+frac{1}{n}}=1+frac{1}{2n}+o(1/n)$.






              share|cite|improve this answer











              $endgroup$



              One more approach. By Stolz-Cesaro Theorem
              $$begin{align}lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}}{nsqrt{n}} &= lim_{n to infty} frac{sqrt{n}}{(n+1)sqrt{n+1}-nsqrt{n}}
              \&=lim_{n to infty} frac{1}{(n+1)sqrt{1+frac{1}{n}}-n}\
              &=frac{1}{1+frac{1}{2}}=frac{2}{3}end{align}$$

              where we used the fact that $sqrt{1+frac{1}{n}}=1+frac{1}{2n}+o(1/n)$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 15 '18 at 17:28

























              answered Dec 15 '18 at 9:45









              Robert ZRobert Z

              99.8k1068140




              99.8k1068140












              • $begingroup$
                But I do not understand how this Stolz-Cesaro Theorem would help me get the $dfrac{sqrt{n}}{(n+1){sqrt{n+1}}-nsqrt{n}}$ .Please elaborate because according to this theorem if $lim_{n to infty}dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}=l implies lim_{n to infty} dfrac{a_n}{b_n}=l $.But how does this relate to the fraction you obtained. Please dont mind my stupidity for not understanding this..
                $endgroup$
                – Saradamani
                Dec 15 '18 at 16:35












              • $begingroup$
                @Saradamani I added a few details. Is it clear now?
                $endgroup$
                – Robert Z
                Dec 15 '18 at 17:29






              • 1




                $begingroup$
                @Saradamani For Stolz-Cesaro we have $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ and $b_n=nsqrt{n}$.
                $endgroup$
                – Robert Z
                Dec 15 '18 at 17:31






              • 1




                $begingroup$
                No, If $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ then $a_{n+1}-a_n=(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}+sqrt{n})-(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1})=sqrt{n}$
                $endgroup$
                – Robert Z
                Dec 16 '18 at 6:38






              • 1




                $begingroup$
                @Saradamani As a matter of fact, the classic proof of Stolz-Cesaro Theorem (see link above) is based on the sandwich principle...
                $endgroup$
                – Robert Z
                Dec 16 '18 at 7:05




















              • $begingroup$
                But I do not understand how this Stolz-Cesaro Theorem would help me get the $dfrac{sqrt{n}}{(n+1){sqrt{n+1}}-nsqrt{n}}$ .Please elaborate because according to this theorem if $lim_{n to infty}dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}=l implies lim_{n to infty} dfrac{a_n}{b_n}=l $.But how does this relate to the fraction you obtained. Please dont mind my stupidity for not understanding this..
                $endgroup$
                – Saradamani
                Dec 15 '18 at 16:35












              • $begingroup$
                @Saradamani I added a few details. Is it clear now?
                $endgroup$
                – Robert Z
                Dec 15 '18 at 17:29






              • 1




                $begingroup$
                @Saradamani For Stolz-Cesaro we have $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ and $b_n=nsqrt{n}$.
                $endgroup$
                – Robert Z
                Dec 15 '18 at 17:31






              • 1




                $begingroup$
                No, If $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ then $a_{n+1}-a_n=(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}+sqrt{n})-(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1})=sqrt{n}$
                $endgroup$
                – Robert Z
                Dec 16 '18 at 6:38






              • 1




                $begingroup$
                @Saradamani As a matter of fact, the classic proof of Stolz-Cesaro Theorem (see link above) is based on the sandwich principle...
                $endgroup$
                – Robert Z
                Dec 16 '18 at 7:05


















              $begingroup$
              But I do not understand how this Stolz-Cesaro Theorem would help me get the $dfrac{sqrt{n}}{(n+1){sqrt{n+1}}-nsqrt{n}}$ .Please elaborate because according to this theorem if $lim_{n to infty}dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}=l implies lim_{n to infty} dfrac{a_n}{b_n}=l $.But how does this relate to the fraction you obtained. Please dont mind my stupidity for not understanding this..
              $endgroup$
              – Saradamani
              Dec 15 '18 at 16:35






              $begingroup$
              But I do not understand how this Stolz-Cesaro Theorem would help me get the $dfrac{sqrt{n}}{(n+1){sqrt{n+1}}-nsqrt{n}}$ .Please elaborate because according to this theorem if $lim_{n to infty}dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}=l implies lim_{n to infty} dfrac{a_n}{b_n}=l $.But how does this relate to the fraction you obtained. Please dont mind my stupidity for not understanding this..
              $endgroup$
              – Saradamani
              Dec 15 '18 at 16:35














              $begingroup$
              @Saradamani I added a few details. Is it clear now?
              $endgroup$
              – Robert Z
              Dec 15 '18 at 17:29




              $begingroup$
              @Saradamani I added a few details. Is it clear now?
              $endgroup$
              – Robert Z
              Dec 15 '18 at 17:29




              1




              1




              $begingroup$
              @Saradamani For Stolz-Cesaro we have $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ and $b_n=nsqrt{n}$.
              $endgroup$
              – Robert Z
              Dec 15 '18 at 17:31




              $begingroup$
              @Saradamani For Stolz-Cesaro we have $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ and $b_n=nsqrt{n}$.
              $endgroup$
              – Robert Z
              Dec 15 '18 at 17:31




              1




              1




              $begingroup$
              No, If $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ then $a_{n+1}-a_n=(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}+sqrt{n})-(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1})=sqrt{n}$
              $endgroup$
              – Robert Z
              Dec 16 '18 at 6:38




              $begingroup$
              No, If $a_n=sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}$ then $a_{n+1}-a_n=(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1}+sqrt{n})-(sqrt{1}+sqrt{2}+sqrt{3}+dots+sqrt{n-1})=sqrt{n}$
              $endgroup$
              – Robert Z
              Dec 16 '18 at 6:38




              1




              1




              $begingroup$
              @Saradamani As a matter of fact, the classic proof of Stolz-Cesaro Theorem (see link above) is based on the sandwich principle...
              $endgroup$
              – Robert Z
              Dec 16 '18 at 7:05






              $begingroup$
              @Saradamani As a matter of fact, the classic proof of Stolz-Cesaro Theorem (see link above) is based on the sandwich principle...
              $endgroup$
              – Robert Z
              Dec 16 '18 at 7:05













              0












              $begingroup$

              It is sufficient to obtain integral of $sqrt{x}$, when $0leq xleq 1$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                It is sufficient to obtain integral of $sqrt{x}$, when $0leq xleq 1$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  It is sufficient to obtain integral of $sqrt{x}$, when $0leq xleq 1$.






                  share|cite|improve this answer









                  $endgroup$



                  It is sufficient to obtain integral of $sqrt{x}$, when $0leq xleq 1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 8:51









                  user479859user479859

                  887




                  887






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3040278%2fvalue-of-lim-n-to-infty-dfrac-sqrt1-sqrt2-sqrt3-sqrtn-1n%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Plaza Victoria

                      Puebla de Zaragoza

                      Musa