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I have a sequence $$a_n=\sum_{k=1}^n\frac n{n^2+k}$$ I've to find out if this sequence is convergent. I've made some progress as to find out if this is bounded. $$\frac n{n^2+1}<a_n<\frac n{n^2+n}$$ so this gives me that $a_n$ is bounded between 0 and 1.

But I am not able to prove if it's convergent. I've read so much that it's really confusing me now. At some places it's saying : if the sequence converges to a limit then its convergent, at other places, it's saying since $a_n\to1$ as $n\to\infty$ then it's not convergent. How to test for convergence in this case, any ideas? Also it would be great if you could point out a definite condition for testing convergence of a sequence.

Martin Sleziak
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slick
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    The fact that $a_n\rightarrow 1$ as $n\rightarrow\infty$ does not mean it's not convergent, maybe what you read was looking at sums like $\sum\limits_{k=1}^n a_k$, in which case we would require $a_k\rightarrow 0$, but you are defining $a_n$ to be the $n$th partial sum, which doesn't have to go to $0$ to converge. – Jürgen Sukumaran Jan 31 '18 at 12:27
  • @TonyS.F. If $a_n \underset{n \rightarrow +\infty}{\rightarrow}1$ then the series $\displaystyle \sum_{ n \geq 1}^{ }a_n$ cannot be convergent. – Atmos Jan 31 '18 at 12:30
  • @TonyS.F. why should the conditions differ for the above two cases you mentioned ? – slick Jan 31 '18 at 12:31
  • @Atmos I agree, but here we are not looking at the sum $\sum\limits_{n\geq 1}a_n$, we are looking at $a_n$ as the $n$th partial sum of some other summand, $a_n=\sum\limits_{k=1}^nb_n$. The difference is whether $a_n$ is the $n$th partial sum or whether $a_n$ is the summand itself. – Jürgen Sukumaran Jan 31 '18 at 12:37
  • @Atmos : The term "series" shouldn't be applied : it is not the partial sums of a series $\sum_{k=1}^{\infty}a_k$, thus one cannot apply results on series ! – Jean Marie Jan 31 '18 at 12:38
  • @slick I think the confusion here is that you are using $a_n$ to be the $n$th partial sum and also $a_n$ to represent the summand $\frac{n}{n^2}+k$, they are not the same. – Jürgen Sukumaran Jan 31 '18 at 12:41
  • Ive to find if the sequence {an} is just bounded or convergent. So the an i've provided is just the nth term of the sequence. – slick Jan 31 '18 at 12:48
  • The answer that I have in my text says the sequence is bounded but not convergent. I really dont see how or what it means! – slick Jan 31 '18 at 12:54
  • It's obviously monotone so if you prove it is bounded then by the monotone convergence theorem we are done. – Jürgen Sukumaran Jan 31 '18 at 12:56

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For $1\le k\le n$ we have:$$\frac{1}{n+1}\le\dfrac{n}{n^2+k}\le\dfrac{n}{n^2+1}$$therefore$$\sum_{k=1}^{n}\dfrac{1}{n+1}\le a_n=\sum_{k=1}^{n}\dfrac{n}{n^2+k}\le\sum_{k=1}^{n}\dfrac{n}{n^2+1}$$or$$\dfrac{n}{n+1}\le a_n\le\dfrac{n^2}{n^2+1}$$which is obviously convergent to 1.

Guy Fsone
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Mostafa Ayaz
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