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Let $ a_{n} = \sum_{k=1}^{n} \frac{n}{n^{2}+k}$ . I would like to know whether the given sequence converges.

I see that,

$ a_{n} = \sum_{k=1}^{n} \frac{n}{n^{2}+k}= \sum_{k=1}^{n} \frac{1}{n+\frac{k}{n}}.$ When $n$ gets sufficiently large the contribution by the $ \frac{k}{n} $ term is diminishing and $ a_{n} < \sum_{k=1}^{n} \frac{1}{n} = 1 $.

Thank you.

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Also note that $a_n>\frac{n}{n+1}$. What does that say about the convergence? –  Thomas Andrews Nov 1 '11 at 17:41
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A sloppy approach seems to yield consistent results: $$\sum\limits_{k=1}^n \frac{n}{n^2+k}=n\left(\sum\limits_{k=1}^{n^2+n}\frac1{k}-\sum\limits_{k=1}^{n^2‌​}\frac1{k}\right)\approx n(\log(n^2+n)-\log(n^2))=\log\left(1+\frac1{n}\right)^n$$... –  J. M. Nov 1 '11 at 17:48
    
$\sum\limits_{k=1}^n \frac{n}{n^2+k}=n(H_{n^2+n}-H_{n^2})$ ,definition of $H$ –  pedja Nov 1 '11 at 17:55
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Thank you all .I found your discussions to be very useful! –  G.Dinesh Nathan Nov 1 '11 at 18:08
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1 Answer 1

up vote 5 down vote accepted

$$\frac{1}{n+1}\leq \frac{1}{n+\frac{k}n} \leq \frac{1}{n}$$

So $\frac{n}{n+1} \leq a_n \leq 1$

So $a_n\rightarrow 1$.

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