Value of $lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+…+sqrt{n-1}}{nsqrt{n}}$
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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
$endgroup$
add a comment |
$begingroup$
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
$limlimits_{n to infty } frac1{sqrt n}left(1 + frac1{sqrt 2 }+frac1{sqrt 3 }+cdots+frac1{sqrt n } right)=limlimits_{n to infty }sum_{k=1}^{n}{dfrac{1}{sqrt{kn}}}=int_{0}^{1}dfrac{1}{sqrt{x}}dx+limlimits_{n to infty }dfrac{1}{n}=2$
calculus limits
$endgroup$
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
add a comment |
$begingroup$
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
$limlimits_{n to infty } frac1{sqrt n}left(1 + frac1{sqrt 2 }+frac1{sqrt 3 }+cdots+frac1{sqrt n } right)=limlimits_{n to infty }sum_{k=1}^{n}{dfrac{1}{sqrt{kn}}}=int_{0}^{1}dfrac{1}{sqrt{x}}dx+limlimits_{n to infty }dfrac{1}{n}=2$
calculus limits
$endgroup$
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
calculus limits
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
add a comment |
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
add a comment |
4 Answers
4
active
oldest
votes
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Hint :
$$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}} = lim_{n to infty} frac{1}{n} sum_{k=0}^{n-1} sqrt{frac{k}{n}} = int_0^1sqrt{x}mathrm{d}x$$
$endgroup$
add a comment |
$begingroup$
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 $a_{10}approx 0.610509$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $sqrt{1+frac{1}{n}}=1+frac{1}{2n}+o(1/n)$.
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$begingroup$
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
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@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
|
show 5 more comments
$begingroup$
It is sufficient to obtain integral of $sqrt{x}$, when $0leq xleq 1$.
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add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint :
$$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}} = lim_{n to infty} frac{1}{n} sum_{k=0}^{n-1} sqrt{frac{k}{n}} = int_0^1sqrt{x}mathrm{d}x$$
$endgroup$
add a comment |
$begingroup$
Hint :
$$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}} = lim_{n to infty} frac{1}{n} sum_{k=0}^{n-1} sqrt{frac{k}{n}} = int_0^1sqrt{x}mathrm{d}x$$
$endgroup$
add a comment |
$begingroup$
Hint :
$$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}} = lim_{n to infty} frac{1}{n} sum_{k=0}^{n-1} sqrt{frac{k}{n}} = int_0^1sqrt{x}mathrm{d}x$$
$endgroup$
Hint :
$$lim_{n to infty}dfrac{sqrt{1}+sqrt{2}+sqrt{3}+...+sqrt{n-1}}{nsqrt{n}} = lim_{n to infty} frac{1}{n} sum_{k=0}^{n-1} sqrt{frac{k}{n}} = int_0^1sqrt{x}mathrm{d}x$$
answered Dec 15 '18 at 8:51
RebellosRebellos
15.1k31250
15.1k31250
add a comment |
add a comment |
$begingroup$
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 $a_{10}approx 0.610509$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $a_{10}approx 0.610509$.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $a_{10}approx 0.610509$.
$endgroup$
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 $a_{10}approx 0.610509$.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $sqrt{1+frac{1}{n}}=1+frac{1}{2n}+o(1/n)$.
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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 $sqrt{1+frac{1}{n}}=1+frac{1}{2n}+o(1/n)$.
$endgroup$
$begingroup$
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
|
show 5 more comments
$begingroup$
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 $sqrt{1+frac{1}{n}}=1+frac{1}{2n}+o(1/n)$.
$endgroup$
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 $sqrt{1+frac{1}{n}}=1+frac{1}{2n}+o(1/n)$.
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
|
show 5 more comments
$begingroup$
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
|
show 5 more comments
$begingroup$
It is sufficient to obtain integral of $sqrt{x}$, when $0leq xleq 1$.
$endgroup$
add a comment |
$begingroup$
It is sufficient to obtain integral of $sqrt{x}$, when $0leq xleq 1$.
$endgroup$
add a comment |
$begingroup$
It is sufficient to obtain integral of $sqrt{x}$, when $0leq xleq 1$.
$endgroup$
It is sufficient to obtain integral of $sqrt{x}$, when $0leq xleq 1$.
answered Dec 15 '18 at 8:51
user479859user479859
887
887
add a comment |
add a comment |
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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
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– 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
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There are $n-1$ terms in the numerator
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– Shubham Johri
Dec 15 '18 at 8:51