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Please help me solve this. Please try to give some details also.

Let $a_n=\sum_{k=1}^n \frac{n}{n^2+k}\ $for $ n \in\mathbb{N}$ then the sequence $\{a_n\}$ is (choose the correct option):

  1. Convergent
  2. Bounded but not convergent
  3. Diverges to infinity
  4. Neither bounded nor diverges to infinity

My thoughts :
I tried doing it by testing if it is monotonic with help of $a_{n+1} -a_n$ and checking if its +ve or -ve but, the the terms dont simplyfy by cancelling each other...

Simply Beautiful Art
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ketan
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  • Welcome to Math.SE! This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Nick Peterson Jan 19 '14 at 16:42

3 Answers3

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Hint

Prove these inequalities $$b_n=n\times \frac{n}{n^2+n}\le a_n\le n\times \frac{n}{n^2+1}=c_n$$ What's the limit of $(b_n)$ and $(c_n)$ and conclude.

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$\begin{array}\\ d_n &=a_n-1\\ &=\sum_{k=1}^n \frac{n}{n^2+k}-1\\ &=\sum_{k=1}^n \frac{n}{n^2+k}-\sum_{k=1}^n\frac1{n}\\ &=\sum_{k=1}^n (\frac{n}{n^2+k}-\frac1{n})\\ &=\sum_{k=1}^n \frac{n^2-(n^2+k)}{n(n^2+k)}\\ &=\sum_{k=1}^n \frac{-k}{n(n^2+k)}\\ &=-\sum_{k=1}^n \frac{k}{n(n^2+k)}\\ \text{so}\\ d_n &>-\sum_{k=1}^n \frac{k}{n(n^2)}\\ &=-\frac{n(n+1)}{2n^3}\\ &=-\frac{n+1}{2n^2}\\ &=-\frac{1+1/n}{2n}\\ d_n &<-\sum_{k=1}^n \frac{k}{n(n^2+n)}\\ &=-\frac{n(n+1)}{2n(n^2+n)}\\ &=-\frac{1}{2n}\\ \end{array} $

So

$\frac{1}{2n} \lt 1-a_n \lt \frac{1+1/n}{2n} = \frac{1}{2n}+\frac1{2n^2} $.

From this, $a_n \to 1$.

Even more previsely, $0 \lt n(1-a_n)-\frac12 \lt\frac1{2n} $ so $n(1-a_n) \to \frac12 $.

marty cohen
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$$ \frac{1}{n+1}\le\frac{n}{n^2+k}\le\frac{1}{n}, $$ and thus $$ 1-\frac{1}{n+1}=\frac{n}{n+1}=\sum_{k=1}^n\frac{n}{n^2+k}\le n\cdot\frac{1}{n}=1. $$ Hence only 1. is valid, and the sequence converges to $1$.

Yiorgos S. Smyrlis
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  • is there some reason that motivates to approach this particular problem in this way using inequalities? i mean books too solve using inequality just as urs, do you have a concepual reason , why to approach this problem like that only? – ketan Jan 19 '14 at 17:09
  • @ketan: I am using the fact that if $a_n\le b_n\le c_n$ and $\lim_{n\to\infty}a_n=\lim_{n\to\infty}c_n=\ell$, then $b_n$ also converges and its limit is $\ell$. – Yiorgos S. Smyrlis Jan 19 '14 at 17:14
  • oh sandwich i see, but is your inequality true without summation symbol? isn't it false without sigma? – ketan Jan 19 '14 at 17:19