Find $\lim\limits_{n\to\infty} ( \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+ \frac{1}{(2n)^2})$

Question

Please help me find this limit .

I cannot use cauchys limit theorem on this problem . So I am stuck and clueless

Related : http://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2 — lab bhattacharjee, Jan 24 '17 at 05:39

score 3 · Accepted Answer · answered Jan 24 '17 at 04:11

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\sum_{k = 1}^{n}{1 \over \pars{k + n}^{2}} = \lim_{n \to \infty}\braces{{1 \over n} \bracks{{1 \over n}\sum_{k = 1}^{n}{1 \over \pars{k/n + 1}^{2}}}} \end{align} Note that $\ds{{1 \over n}\sum_{k = 1}^{n}{1 \over \pars{k/n + 1}^{2}}}$ is a convergent Riemann sum $\ds{\pars{~\mbox{as}\ n \to \infty~}}$ such that the above limit $\underline{\mbox{vanishes out}}$.

@AbhishekChandra You're welcome. – Felix Marin Jan 24 '17 at 20:35 — Felix Marin, Jan 24 '17 at 20:35

score 3 · Answer 2 · answered Jan 24 '17 at 09:33

3

Observe that

$$0 < \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+ \frac{1}{(2n)^2} < n\cdot\frac{1}{(n+1)^2}.$$

Since the limit of the right side is $0,$ the squeeze theorem shows our limit is $0.$

answered Jan 24 '17 at 09:33

zhw.

105,693

score 2 · Answer 3 · answered Jan 24 '17 at 03:14

2

We know that $\sum\limits_{n=1}^\infty \frac{1}{n^2}$ converges. It follows that $\lim\limits_{m\to \infty}\sum\limits_{n=m+1}^\infty\frac{1}{n^2}=0$.

Notice that $0\leq \sum\limits_{n=m+1}^{2m}\frac{1}{n^2}\leq \sum\limits_{n=m+1}^\infty \frac{1}{n^2}$. Now just use the squeeze theorem.

answered Jan 24 '17 at 03:14

Asinomás

105,651

Can you give me a solution without sandwich theorem – Abhishek Chandra Jan 24 '17 at 03:15
Actually I am not familiar with sandwich theorem..A solution with Cauchy's theorem will be helpful – Abhishek Chandra Jan 24 '17 at 03:15
https://www.khanacademy.org/math/ap-calculus-ab/limits-from-equations-ab/squeeze-theorem-ab/v/squeeze-sandwich-theorem you can learn about it here, it is quite useful. – Asinomás Jan 24 '17 at 03:16
Thanks for that..But just for my interest can it be solvable by other method that will be very kind of you – Abhishek Chandra Jan 24 '17 at 03:30

score 1 · Answer 4 · answered Jan 24 '17 at 07:24

For $j\in \mathbb N$ and $x\in (j,j+1)$ we have $\frac {1}{(j+1)^2}<\frac {1}{x^2}$. Therefore $$0<\sum_{j=n}^{2n-1}\frac {1}{(j+1)^2}=\sum_{j=n}^{2n-1}\int_j^{j+1}\frac {1}{(j+1)^2}dx<\sum_{j=n}^{2n-1}\int_j^{j+1} \frac {1}{x^2}dx=$$ $$=\int_n^{2n}\frac {1}{x^2}dx=\frac {1}{n}- \frac {1}{2n}=\frac {1}{2n}.$$

Find $\lim\limits_{n\to\infty} ( \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+ \frac{1}{(2n)^2})$

4 Answers4