How to check if the sum of infinite series is convergent?












4












$begingroup$


I have this exercise where I need to find if the sum of infinite series is convergent:



$sum_{n=1}^ infty frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)} $
for x $ in (pi/2,pi) $



Now I decided to do a ratio test for $ frac{|a_n+1|}{a_n} $ but I am currently stuck on simplifying the result and proceeding with the solution.



$ frac{(sin^2(x) - sin (x) +1)^{n+1}}{ln(2+n)} cdot frac{ln(1+n)}{(sin^2(x) - sin (x) +1)^{n}}$



I am not really sure how to proceed from here, any further help would be appreciated, thanks!










share|cite|improve this question











$endgroup$












  • $begingroup$
    $0<sin^2x-sin x+1<1$ in that interval.
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:43












  • $begingroup$
    why specifically $sin^2 x - sin x +1$ ? also what about the natural log? and should I be checking for $ frac{|a_n+1|}{a_n}$?
    $endgroup$
    – Mark
    Dec 11 '18 at 21:45












  • $begingroup$
    For any $x$, the numerator is equal to $r^n$ where $0<r<1$. For $n>1$, the denominator is greater than $1$. So your series is less-than-or-equal-to $sum r^n$
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:50


















4












$begingroup$


I have this exercise where I need to find if the sum of infinite series is convergent:



$sum_{n=1}^ infty frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)} $
for x $ in (pi/2,pi) $



Now I decided to do a ratio test for $ frac{|a_n+1|}{a_n} $ but I am currently stuck on simplifying the result and proceeding with the solution.



$ frac{(sin^2(x) - sin (x) +1)^{n+1}}{ln(2+n)} cdot frac{ln(1+n)}{(sin^2(x) - sin (x) +1)^{n}}$



I am not really sure how to proceed from here, any further help would be appreciated, thanks!










share|cite|improve this question











$endgroup$












  • $begingroup$
    $0<sin^2x-sin x+1<1$ in that interval.
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:43












  • $begingroup$
    why specifically $sin^2 x - sin x +1$ ? also what about the natural log? and should I be checking for $ frac{|a_n+1|}{a_n}$?
    $endgroup$
    – Mark
    Dec 11 '18 at 21:45












  • $begingroup$
    For any $x$, the numerator is equal to $r^n$ where $0<r<1$. For $n>1$, the denominator is greater than $1$. So your series is less-than-or-equal-to $sum r^n$
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:50
















4












4








4





$begingroup$


I have this exercise where I need to find if the sum of infinite series is convergent:



$sum_{n=1}^ infty frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)} $
for x $ in (pi/2,pi) $



Now I decided to do a ratio test for $ frac{|a_n+1|}{a_n} $ but I am currently stuck on simplifying the result and proceeding with the solution.



$ frac{(sin^2(x) - sin (x) +1)^{n+1}}{ln(2+n)} cdot frac{ln(1+n)}{(sin^2(x) - sin (x) +1)^{n}}$



I am not really sure how to proceed from here, any further help would be appreciated, thanks!










share|cite|improve this question











$endgroup$




I have this exercise where I need to find if the sum of infinite series is convergent:



$sum_{n=1}^ infty frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)} $
for x $ in (pi/2,pi) $



Now I decided to do a ratio test for $ frac{|a_n+1|}{a_n} $ but I am currently stuck on simplifying the result and proceeding with the solution.



$ frac{(sin^2(x) - sin (x) +1)^{n+1}}{ln(2+n)} cdot frac{ln(1+n)}{(sin^2(x) - sin (x) +1)^{n}}$



I am not really sure how to proceed from here, any further help would be appreciated, thanks!







calculus sequences-and-series convergence trigonometric-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 22:43







Mark

















asked Dec 11 '18 at 21:37









MarkMark

233




233












  • $begingroup$
    $0<sin^2x-sin x+1<1$ in that interval.
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:43












  • $begingroup$
    why specifically $sin^2 x - sin x +1$ ? also what about the natural log? and should I be checking for $ frac{|a_n+1|}{a_n}$?
    $endgroup$
    – Mark
    Dec 11 '18 at 21:45












  • $begingroup$
    For any $x$, the numerator is equal to $r^n$ where $0<r<1$. For $n>1$, the denominator is greater than $1$. So your series is less-than-or-equal-to $sum r^n$
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:50




















  • $begingroup$
    $0<sin^2x-sin x+1<1$ in that interval.
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:43












  • $begingroup$
    why specifically $sin^2 x - sin x +1$ ? also what about the natural log? and should I be checking for $ frac{|a_n+1|}{a_n}$?
    $endgroup$
    – Mark
    Dec 11 '18 at 21:45












  • $begingroup$
    For any $x$, the numerator is equal to $r^n$ where $0<r<1$. For $n>1$, the denominator is greater than $1$. So your series is less-than-or-equal-to $sum r^n$
    $endgroup$
    – David Peterson
    Dec 11 '18 at 21:50


















$begingroup$
$0<sin^2x-sin x+1<1$ in that interval.
$endgroup$
– David Peterson
Dec 11 '18 at 21:43






$begingroup$
$0<sin^2x-sin x+1<1$ in that interval.
$endgroup$
– David Peterson
Dec 11 '18 at 21:43














$begingroup$
why specifically $sin^2 x - sin x +1$ ? also what about the natural log? and should I be checking for $ frac{|a_n+1|}{a_n}$?
$endgroup$
– Mark
Dec 11 '18 at 21:45






$begingroup$
why specifically $sin^2 x - sin x +1$ ? also what about the natural log? and should I be checking for $ frac{|a_n+1|}{a_n}$?
$endgroup$
– Mark
Dec 11 '18 at 21:45














$begingroup$
For any $x$, the numerator is equal to $r^n$ where $0<r<1$. For $n>1$, the denominator is greater than $1$. So your series is less-than-or-equal-to $sum r^n$
$endgroup$
– David Peterson
Dec 11 '18 at 21:50






$begingroup$
For any $x$, the numerator is equal to $r^n$ where $0<r<1$. For $n>1$, the denominator is greater than $1$. So your series is less-than-or-equal-to $sum r^n$
$endgroup$
– David Peterson
Dec 11 '18 at 21:50












4 Answers
4






active

oldest

votes


















2












$begingroup$

hint



$$ln(n+2)=lnBigl((n+1)(1+frac{1}{n+1})Bigr)$$



The limit of the ratio is



$$L=sin^2(x)-sin(x)+1$$



but



$$-1<sin(x)Bigl(sin(x)-1Bigr)<0$$



thus



$$0<L<1$$
the series is convergent.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
    $endgroup$
    – Mark
    Dec 11 '18 at 21:49










  • $begingroup$
    @Mark The ratio of $ln$ goes to one.
    $endgroup$
    – hamam_Abdallah
    Dec 11 '18 at 21:50



















2












$begingroup$

Show that if $0<u<1,$ then $0<u^2-u+1<1.$ Now fix $xin (pi/2,pi).$ Then $0<sin x <1.$ From the above, $0<sin^2 x - sin x + 1 <1.$ Hence



$$sum_{n=1}^{infty} (sin^2 x - sin x + 1)^n$$



is a positive convergent series. Dividing by $ln (n+1)$ only helps. Thus your series converges for all $xin (pi/2,pi).$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
    $endgroup$
    – Mark
    Dec 11 '18 at 22:48












  • $begingroup$
    Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
    $endgroup$
    – zhw.
    Dec 11 '18 at 22:57












  • $begingroup$
    (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
    $endgroup$
    – Mark Viola
    Dec 11 '18 at 23:09












  • $begingroup$
    @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
    $endgroup$
    – zhw.
    Dec 11 '18 at 23:16










  • $begingroup$
    @zhw. Happy Holiday to you Z-man!
    $endgroup$
    – Mark Viola
    Dec 12 '18 at 2:27



















1












$begingroup$

As an alternative by root test



$$sqrt[n]{frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)}}tosin^2(x) - sin (x) +1$$



and on the interval



$$0<sin^2(x) - sin (x) +1<1$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    PARTIAL PROOF



    Find the radius of convergence of the series
    $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(ln left(n+1right)right)}{lnleft(n+2right)}left(sin x^2-sin x+1right)right|$



    As n tends to ∞,
    Using L'Hospital's Rule,



    $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(left(n+2right)right)}{n+1}left(sin x^2-sin x+1right)right|$



    =$left|left(sin x^2-sin x+1right)right|$ (As n tends to ∞)



    This must be <1 for the series to converge



    Therefore,



    $left|left(sin x^2-sin x+1right)right|<1$ for the series to converge



    Prove that the interval lies in these values.






    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%2f3035864%2fhow-to-check-if-the-sum-of-infinite-series-is-convergent%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









      2












      $begingroup$

      hint



      $$ln(n+2)=lnBigl((n+1)(1+frac{1}{n+1})Bigr)$$



      The limit of the ratio is



      $$L=sin^2(x)-sin(x)+1$$



      but



      $$-1<sin(x)Bigl(sin(x)-1Bigr)<0$$



      thus



      $$0<L<1$$
      the series is convergent.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
        $endgroup$
        – Mark
        Dec 11 '18 at 21:49










      • $begingroup$
        @Mark The ratio of $ln$ goes to one.
        $endgroup$
        – hamam_Abdallah
        Dec 11 '18 at 21:50
















      2












      $begingroup$

      hint



      $$ln(n+2)=lnBigl((n+1)(1+frac{1}{n+1})Bigr)$$



      The limit of the ratio is



      $$L=sin^2(x)-sin(x)+1$$



      but



      $$-1<sin(x)Bigl(sin(x)-1Bigr)<0$$



      thus



      $$0<L<1$$
      the series is convergent.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
        $endgroup$
        – Mark
        Dec 11 '18 at 21:49










      • $begingroup$
        @Mark The ratio of $ln$ goes to one.
        $endgroup$
        – hamam_Abdallah
        Dec 11 '18 at 21:50














      2












      2








      2





      $begingroup$

      hint



      $$ln(n+2)=lnBigl((n+1)(1+frac{1}{n+1})Bigr)$$



      The limit of the ratio is



      $$L=sin^2(x)-sin(x)+1$$



      but



      $$-1<sin(x)Bigl(sin(x)-1Bigr)<0$$



      thus



      $$0<L<1$$
      the series is convergent.






      share|cite|improve this answer









      $endgroup$



      hint



      $$ln(n+2)=lnBigl((n+1)(1+frac{1}{n+1})Bigr)$$



      The limit of the ratio is



      $$L=sin^2(x)-sin(x)+1$$



      but



      $$-1<sin(x)Bigl(sin(x)-1Bigr)<0$$



      thus



      $$0<L<1$$
      the series is convergent.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 11 '18 at 21:45









      hamam_Abdallahhamam_Abdallah

      38.1k21634




      38.1k21634












      • $begingroup$
        Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
        $endgroup$
        – Mark
        Dec 11 '18 at 21:49










      • $begingroup$
        @Mark The ratio of $ln$ goes to one.
        $endgroup$
        – hamam_Abdallah
        Dec 11 '18 at 21:50


















      • $begingroup$
        Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
        $endgroup$
        – Mark
        Dec 11 '18 at 21:49










      • $begingroup$
        @Mark The ratio of $ln$ goes to one.
        $endgroup$
        – hamam_Abdallah
        Dec 11 '18 at 21:50
















      $begingroup$
      Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
      $endgroup$
      – Mark
      Dec 11 '18 at 21:49




      $begingroup$
      Thanks for the answer, may I ask why is the L just $sin^2(x)−sin(x)+1$ and why are we not checking for the ln?
      $endgroup$
      – Mark
      Dec 11 '18 at 21:49












      $begingroup$
      @Mark The ratio of $ln$ goes to one.
      $endgroup$
      – hamam_Abdallah
      Dec 11 '18 at 21:50




      $begingroup$
      @Mark The ratio of $ln$ goes to one.
      $endgroup$
      – hamam_Abdallah
      Dec 11 '18 at 21:50











      2












      $begingroup$

      Show that if $0<u<1,$ then $0<u^2-u+1<1.$ Now fix $xin (pi/2,pi).$ Then $0<sin x <1.$ From the above, $0<sin^2 x - sin x + 1 <1.$ Hence



      $$sum_{n=1}^{infty} (sin^2 x - sin x + 1)^n$$



      is a positive convergent series. Dividing by $ln (n+1)$ only helps. Thus your series converges for all $xin (pi/2,pi).$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
        $endgroup$
        – Mark
        Dec 11 '18 at 22:48












      • $begingroup$
        Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
        $endgroup$
        – zhw.
        Dec 11 '18 at 22:57












      • $begingroup$
        (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
        $endgroup$
        – Mark Viola
        Dec 11 '18 at 23:09












      • $begingroup$
        @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
        $endgroup$
        – zhw.
        Dec 11 '18 at 23:16










      • $begingroup$
        @zhw. Happy Holiday to you Z-man!
        $endgroup$
        – Mark Viola
        Dec 12 '18 at 2:27
















      2












      $begingroup$

      Show that if $0<u<1,$ then $0<u^2-u+1<1.$ Now fix $xin (pi/2,pi).$ Then $0<sin x <1.$ From the above, $0<sin^2 x - sin x + 1 <1.$ Hence



      $$sum_{n=1}^{infty} (sin^2 x - sin x + 1)^n$$



      is a positive convergent series. Dividing by $ln (n+1)$ only helps. Thus your series converges for all $xin (pi/2,pi).$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
        $endgroup$
        – Mark
        Dec 11 '18 at 22:48












      • $begingroup$
        Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
        $endgroup$
        – zhw.
        Dec 11 '18 at 22:57












      • $begingroup$
        (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
        $endgroup$
        – Mark Viola
        Dec 11 '18 at 23:09












      • $begingroup$
        @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
        $endgroup$
        – zhw.
        Dec 11 '18 at 23:16










      • $begingroup$
        @zhw. Happy Holiday to you Z-man!
        $endgroup$
        – Mark Viola
        Dec 12 '18 at 2:27














      2












      2








      2





      $begingroup$

      Show that if $0<u<1,$ then $0<u^2-u+1<1.$ Now fix $xin (pi/2,pi).$ Then $0<sin x <1.$ From the above, $0<sin^2 x - sin x + 1 <1.$ Hence



      $$sum_{n=1}^{infty} (sin^2 x - sin x + 1)^n$$



      is a positive convergent series. Dividing by $ln (n+1)$ only helps. Thus your series converges for all $xin (pi/2,pi).$






      share|cite|improve this answer









      $endgroup$



      Show that if $0<u<1,$ then $0<u^2-u+1<1.$ Now fix $xin (pi/2,pi).$ Then $0<sin x <1.$ From the above, $0<sin^2 x - sin x + 1 <1.$ Hence



      $$sum_{n=1}^{infty} (sin^2 x - sin x + 1)^n$$



      is a positive convergent series. Dividing by $ln (n+1)$ only helps. Thus your series converges for all $xin (pi/2,pi).$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 11 '18 at 22:08









      zhw.zhw.

      73.6k43175




      73.6k43175












      • $begingroup$
        I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
        $endgroup$
        – Mark
        Dec 11 '18 at 22:48












      • $begingroup$
        Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
        $endgroup$
        – zhw.
        Dec 11 '18 at 22:57












      • $begingroup$
        (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
        $endgroup$
        – Mark Viola
        Dec 11 '18 at 23:09












      • $begingroup$
        @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
        $endgroup$
        – zhw.
        Dec 11 '18 at 23:16










      • $begingroup$
        @zhw. Happy Holiday to you Z-man!
        $endgroup$
        – Mark Viola
        Dec 12 '18 at 2:27


















      • $begingroup$
        I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
        $endgroup$
        – Mark
        Dec 11 '18 at 22:48












      • $begingroup$
        Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
        $endgroup$
        – zhw.
        Dec 11 '18 at 22:57












      • $begingroup$
        (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
        $endgroup$
        – Mark Viola
        Dec 11 '18 at 23:09












      • $begingroup$
        @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
        $endgroup$
        – zhw.
        Dec 11 '18 at 23:16










      • $begingroup$
        @zhw. Happy Holiday to you Z-man!
        $endgroup$
        – Mark Viola
        Dec 12 '18 at 2:27
















      $begingroup$
      I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
      $endgroup$
      – Mark
      Dec 11 '18 at 22:48






      $begingroup$
      I plotted the limit for $sin^2x−sin x+1$ and tried to add the values of π/2 and π, but I get the answer to be equal to 1, why are we saying it is less than 1? Could you please explain this point further?
      $endgroup$
      – Mark
      Dec 11 '18 at 22:48














      $begingroup$
      Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
      $endgroup$
      – zhw.
      Dec 11 '18 at 22:57






      $begingroup$
      Why are you adding the values $pi/2$ and $pi$ when you have told us they are not included?
      $endgroup$
      – zhw.
      Dec 11 '18 at 22:57














      $begingroup$
      (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
      $endgroup$
      – Mark Viola
      Dec 11 '18 at 23:09






      $begingroup$
      (+1) I don't understand why this answer was neither accepted nor upvoted … until now.
      $endgroup$
      – Mark Viola
      Dec 11 '18 at 23:09














      $begingroup$
      @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
      $endgroup$
      – zhw.
      Dec 11 '18 at 23:16




      $begingroup$
      @MarkViola Hey MV, happy holidays! The ways of MSE are indeed mysterious.
      $endgroup$
      – zhw.
      Dec 11 '18 at 23:16












      $begingroup$
      @zhw. Happy Holiday to you Z-man!
      $endgroup$
      – Mark Viola
      Dec 12 '18 at 2:27




      $begingroup$
      @zhw. Happy Holiday to you Z-man!
      $endgroup$
      – Mark Viola
      Dec 12 '18 at 2:27











      1












      $begingroup$

      As an alternative by root test



      $$sqrt[n]{frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)}}tosin^2(x) - sin (x) +1$$



      and on the interval



      $$0<sin^2(x) - sin (x) +1<1$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        As an alternative by root test



        $$sqrt[n]{frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)}}tosin^2(x) - sin (x) +1$$



        and on the interval



        $$0<sin^2(x) - sin (x) +1<1$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          As an alternative by root test



          $$sqrt[n]{frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)}}tosin^2(x) - sin (x) +1$$



          and on the interval



          $$0<sin^2(x) - sin (x) +1<1$$






          share|cite|improve this answer









          $endgroup$



          As an alternative by root test



          $$sqrt[n]{frac{(sin^2(x) - sin (x) +1)^n}{ln(1+n)}}tosin^2(x) - sin (x) +1$$



          and on the interval



          $$0<sin^2(x) - sin (x) +1<1$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 11 '18 at 21:50









          gimusigimusi

          92.9k84494




          92.9k84494























              0












              $begingroup$

              PARTIAL PROOF



              Find the radius of convergence of the series
              $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(ln left(n+1right)right)}{lnleft(n+2right)}left(sin x^2-sin x+1right)right|$



              As n tends to ∞,
              Using L'Hospital's Rule,



              $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(left(n+2right)right)}{n+1}left(sin x^2-sin x+1right)right|$



              =$left|left(sin x^2-sin x+1right)right|$ (As n tends to ∞)



              This must be <1 for the series to converge



              Therefore,



              $left|left(sin x^2-sin x+1right)right|<1$ for the series to converge



              Prove that the interval lies in these values.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                PARTIAL PROOF



                Find the radius of convergence of the series
                $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(ln left(n+1right)right)}{lnleft(n+2right)}left(sin x^2-sin x+1right)right|$



                As n tends to ∞,
                Using L'Hospital's Rule,



                $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(left(n+2right)right)}{n+1}left(sin x^2-sin x+1right)right|$



                =$left|left(sin x^2-sin x+1right)right|$ (As n tends to ∞)



                This must be <1 for the series to converge



                Therefore,



                $left|left(sin x^2-sin x+1right)right|<1$ for the series to converge



                Prove that the interval lies in these values.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  PARTIAL PROOF



                  Find the radius of convergence of the series
                  $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(ln left(n+1right)right)}{lnleft(n+2right)}left(sin x^2-sin x+1right)right|$



                  As n tends to ∞,
                  Using L'Hospital's Rule,



                  $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(left(n+2right)right)}{n+1}left(sin x^2-sin x+1right)right|$



                  =$left|left(sin x^2-sin x+1right)right|$ (As n tends to ∞)



                  This must be <1 for the series to converge



                  Therefore,



                  $left|left(sin x^2-sin x+1right)right|<1$ for the series to converge



                  Prove that the interval lies in these values.






                  share|cite|improve this answer









                  $endgroup$



                  PARTIAL PROOF



                  Find the radius of convergence of the series
                  $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(ln left(n+1right)right)}{lnleft(n+2right)}left(sin x^2-sin x+1right)right|$



                  As n tends to ∞,
                  Using L'Hospital's Rule,



                  $frac{|a_{N+1}|}{|a_N|}$=$left|frac{left(left(n+2right)right)}{n+1}left(sin x^2-sin x+1right)right|$



                  =$left|left(sin x^2-sin x+1right)right|$ (As n tends to ∞)



                  This must be <1 for the series to converge



                  Therefore,



                  $left|left(sin x^2-sin x+1right)right|<1$ for the series to converge



                  Prove that the interval lies in these values.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 11 '18 at 22:07









                  Pratik ApshingePratik Apshinge

                  305




                  305






























                      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%2f3035864%2fhow-to-check-if-the-sum-of-infinite-series-is-convergent%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