# find the point of convergence of sequence {$a_n$} [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 9 at 13:34

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## 2 Answers

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$$\lim_{n\to \infty}a_n=1$$ • Already posted by another user, right? – Did Jan 9 at 15:34