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I would like to find an equivalence at infinity to the following sequence. $$C_n= \sum_{k=1}^{\infty} \frac{n}{nk^2+k+1}$$ Here are the given questions

1-Find the $\lim_{n\to\infty}C_n$

2- If the limit blows up give an equivalence at infinity.

For the first question, I set $$C_n(k)= \frac{n}{nk^2+k+1}$$

then $$\lim_{n\to\infty}C_n(k) =\frac{1}{k^2} $$ and I proved that the sequence $(C_n(k))_n$ is monotone it therefore springs from monotone convergence that, $$\lim_{n\to\infty}C_n=\lim_{n\to\infty}\sum_{k=1}^{\infty} \frac{n}{nk^2+k+1}= \sum_{k=1}^{\infty} \lim_{n\to\infty}\frac{n}{nk^2+k+1} = \sum_{k=1}^{\infty} \frac{1}{k^2} =\frac{\pi^2}{6}$$

Is this reasoning correct if yes then I am interested on knowing some more elegant way (if there is some) of solving this. if not give me answer. I am don't how to solve the question in the case the limit is infinity.

Guy Fsone
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  • Look here: https://math.stackexchange.com/questions/15240/when-can-you-switch-the-order-of-limits – Hector Blandin Nov 22 '17 at 20:54
  • $$\sum_{k=0}^{\infty} \frac{1}{k^2} =\frac{\pi^2}{6}$$ is wrong, but if you started all your sums at $k=1$, it would make sense. And it would be elegant enough without all those many typos. –  Nov 22 '17 at 20:55
  • @ProfessorVector you are right it is up to thanks – Guy Fsone Nov 22 '17 at 21:01

3 Answers3

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$$\sum_{k\geq 0}\frac{1}{(k+a)(k+b)}=\frac{\psi(a)-\psi(b)}{a-b}\qquad\left(\scriptstyle{\psi(x)=\frac{d}{dx}\log\Gamma(x)}\right)$$ hence by factoring $k^2+\frac{k}{n}+\frac{1}{n}$ $$\sum_{k\geq 1}\frac{n}{n k^2+k+1}= \frac{\pi^2}{6}-\frac{\zeta(3)+\zeta(4)}{n}+O\left(\frac{1}{n^2}\right).$$ Needless to say, the first terms of the asymptotic expansion can be found by creative telescoping™ too.

Jack D'Aurizio
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HINT:

The first term of the series is $n$. For the remaining part, note that we have

$$\begin{align} \left|\sum_{k=1}^\infty \left(\frac1{k^2}-\frac{1}{k^2+\frac{k+1}{n}}\right)\right|&=\frac1n\sum_{k=1}^\infty \frac{k+1}{k^2\left(k^2+\frac{k+1}{n}\right)}\\\\&\le \frac2n \sum_{k=1}^\infty \frac{1}{k^3} \end{align}$$

Mark Viola
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  • The OP starts all sums at $k=0$. –  Nov 22 '17 at 20:57
  • @ProfessorVector Yes, I know. This is a "HINT" only. Surely, the OP can remove the offending term, $n$. I've edited to discuss the presence of the first term. – Mark Viola Nov 22 '17 at 21:01
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    I've commented, there was no "cowardly". The ignored $k=0$ was only one reason, I politely didn't mention "rep wh...ing": the "quality" of the question was subterranean, and your answer didn't address that at that time. –  Nov 22 '17 at 21:16
  • In what way is the 'quality' of this question subterranean? The language is perhaps a little awkward but OP has asked a sensible question, given their own reasoning, and asked for it to be checked. I'm not going to disagree that we should have solid standards for questions but this one shows more effort than many of the questions around here. – Steven Stadnicki Nov 22 '17 at 21:20
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    @Steven Stadnicki Those sums starting at $k=0$ are not sensible, and the OP didn't correct that even after being told. Yes, there is own effort, that's why I didn't downvote the question, but somebody who answers should know better, and address the mistakes. –  Nov 22 '17 at 21:32
  • @ProfessorVector if I change to $k=1$ then jack have to update hi result. which is correct – Guy Fsone Nov 22 '17 at 21:45
  • @ProfessorVector I changed now I think is result perfecly feet now – Guy Fsone Nov 22 '17 at 21:46
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From integration point of view, $C_n=\int_{\Bbb N} f(n,k) d m(k)$ with $f(n,k)= \frac{n}{nk^2+k+1}$ where $dm$ is counting measure: m(A)=# A= card (A). We clearly have that $$ \lim_{n\to \infty} f(n,k)= g(k)=\frac{1}{k^2}\quad \text{and}\quad f(n,k)\leq g(k)=\frac{1}{k^2}\in L^1(\Bbb N, dm)$$ since $$\int_{\Bbb N} g(k) d m(k)=\sum_{k\in\Bbb N} \frac{1}{k^2}=\frac{\pi^2}{6}.$$

By convergence dominated theorem we have $$ \lim_{n\to \infty} C_n= \int_{\Bbb N} \lim_{n\to \infty} f(n,k) d m(k)=\int_{\Bbb N}g(k)d m(k)=\sum_{k\in \Bbb N} \frac{1}{k^2}=\frac{\pi^2}{6}$$

Guy Fsone
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