Find $\lim\limits_{n\to\infty} ( \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+ \frac{1}{(2n)^2})$ Find $\lim\limits_{n\to\infty} ( \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+ \frac{1}{(2n)^2})$
Please help me find this limit .
I cannot use cauchys limit theorem on this problem . So I am stuck and clueless
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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 $\underline{\mbox{vanishes out}}$.
A: 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 $0.$
A: 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.
A: For $j\in \mathbb N$ and $x\in (j,j+1)$ we have $\frac {1}{(j+1)^2}<\frac {1}{x^2}$.  Therefore $$0<\sum_{j=n}^{2n-1}\frac {1}{(j+1)^2}=\sum_{j=n}^{2n-1}\int_j^{j+1}\frac {1}{(j+1)^2}dx<\sum_{j=n}^{2n-1}\int_j^{j+1}
\frac {1}{x^2}dx=$$ $$=\int_n^{2n}\frac {1}{x^2}dx=\frac {1}{n}- \frac {1}{2n}=\frac {1}{2n}.$$
