# Limit $\mathop {\lim }\limits_{n \to \infty } n({1 \over {{{(n + 1)}^2}}} + {1 \over {{{(n + 2)}^2}}} + \cdots{1 \over {{{(2n)}^2}}})$

Without using integrals, how to find this limit:

$$\mathop {\lim }\limits_{n \to \infty } {a_n} = n\cdot\left({1 \over {{{(n + 1)}^2}}} + {1 \over {{{(n + 2)}^2}}} + \cdots{1 \over {{{(2n)}^2}}}\right)$$

I tried squeezing the sequence but it didn't workout.
What next should I do?

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The tag (limit-theorems) is not a good fit for this questions, see the tag-wiki. – Martin Sleziak Nov 29 '13 at 8:05

$$\sum_{i=n+1}^{2n}{\frac{1}{i^2}} \leq \sum_{i=n+1}^{2n}{\frac{1}{i(i-1)}}=\sum_{i=n+1}^{2n}{\left(\frac{1}{i-1}-\frac{1}{i}\right)}=\frac{1}{n}-\frac{1}{2n}=\frac{1}{2n}$$

$$\sum_{i=n+1}^{2n}{\frac{1}{i^2}} \geq \sum_{i=n+1}^{2n}{\frac{1}{i(i+1)}}=\sum_{i=n+1}^{2n}{\left(\frac{1}{i}-\frac{1}{i+1}\right)}=\frac{1}{n+1}-\frac{1}{2n+1}=\frac{n}{(n+1)(2n+1)}$$

Now use squeeze theorem to find $\lim_{n \to \infty}{a_n}$.

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telescoping series. nice one! – Daniel Gagnon Nov 27 '13 at 20:54

It's value is equal to $\dfrac{1}{2}$, using definite integral as a limit of sum where $f(x)=\frac{1}{x^2}$.

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