# Limit - Could you help me with it

Can you help me with this limit? What do I have to do? I'm lost.

$$\lim_{n\to\infty}n\left(\sum_{i=1}^{n}\dfrac{1}{(n+i)^2}\right)$$

The solution given is $\dfrac{1}{2}$.

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Note that $$n\left(\sum_{i=1}^{n}\dfrac{1}{(n+i)^2}\right)=\frac{1}{n}\sum_{i=1}^{n}\dfrac{1}{(1+\frac{i}{n})^2}.$$ By Riemann sum, we have $$\lim_{n\to\infty}n\left(\sum_{i=1}^{n}\dfrac{1}{(n+i)^2}\right)=\int_0^1\frac{dx}{(1+x)^2}=-\frac{1}{1+x}\Big|_0^1=-\frac{1}{2}+1=\frac{1}{2}.$$

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If one were being pedantic, isn't this a bit circular, given that integrals are often defined in terms of Riemann Sums? I understand that that would be missing the point (the integral is easier in this case), but it does make this solution a bit unsatisfying, for lack of a better word. – Tim Seguine Jun 16 '14 at 13:35

Paul gave you the nice way for the solution of your problem.

For your curiosity, I shall not enter into much details but I shall just mention that $$\sum_{i=1}^{n}\dfrac{1}{(n+i)^2}$$ has a closed form which has an asymptotic expansion given by $$\frac{1}{2 n}-\frac{3}{8 n^2}+\frac{7}{48 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$

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Compare with the integral of $\frac{n}{x^2}$ from $n$ to $2n$, and manage the error of the approximation.

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Uh? How to manage what "error of approximation"? – DonAntonio Jun 16 '14 at 3:25
Some error control can be done via: $\frac{E(x)}{n}=|\sum_{i=1}^{n} \frac{1}{(n+i)^2} - \int_0^n \frac{1}{(n+x)^2}dx| = |\sum_{i=1}^n ( \frac{1}{(n+i)^2} - \int_{i -1}^{i} \frac{1}{(n+x)^2}dx )| = |\sum_{i=1}^n \int_{ i -1}^i \frac{1}{(n+i)^2} -\frac{1}{(n+x)^2} dx| \leq | \sum_{i=1}^n \frac{1}{(n+i)^2} -\frac{1}{(n+i-1)^2}|=|\frac{1}{(n+1)^2} - \frac{1}{4n^2}|\leq \frac{3}{4n^2}$. – PenasRaul Jun 16 '14 at 3:50