Let $\displaystyle a_n= \sum_{k=1}^{n} \frac{n}{n^2+k}$, for $n\in \mathbb{N}$. Then what is the nature of sequence $\{a_n\}_{n\in\mathbb{N}}$.

I tried using the Cauchy's general principle of converges for a sequence. But I think that this won't help me as because:

$\displaystyle a_{n+p}= \sum_{k=1}^{n+p} \frac{n+p}{{(n+p)}^2+k}$ and $\displaystyle a_n= \sum_{k=1}^{n} \frac{n}{n^2+k}$

And now if I do $a_{n+p}-a_{n}$ then this won't even cancel a single term.

$a_1$ will have one term.

$a_2$ will have two terms, and so on.

But here the first term in $a_2$ is not the term of $a_1$.

and due to this problem I was unable to use any results of convergence of series of positive terms.

Any help/hint will be appreciated.


marked as duplicate by Martin R, Martin Sleziak, Cesareo, DRF, RRL real-analysis Jan 9 at 13:34

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Try sandwiching $a_n$.

$$a_n = \sum_{k=1}^n \frac{n}{n^2 + k} \le \sum_{k=1}^n \frac{1}{n} = 1.$$ $$a_n = \sum_{k=1}^n \frac{n}{n^2 + k} \ge \sum_{k=1}^n \frac{1}{n + 1} = \frac{n}{n+1}.$$


Note that for $1\le k\le n$ :$${n\over n^2+n}\le {n\over n^2+k}\le {n\over n^2+1}$$therefore $${n^2\over n^2+n}\le a_n\le {n^2\over n^2+1}$$and bu squeeze theorem$$\lim_{n\to \infty}a_n=1$$

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  • $\begingroup$ Already posted by another user, right? $\endgroup$ – Did Jan 9 at 15:34

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