Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\dfrac{1}{2}$.

share|improve this question
3  
What have you tried? –  vonbrand Jun 16 at 2:26

3 Answers 3

up vote 13 down vote accepted

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 $$\lim_{n\to\infty}n\left(\sum_{i=1}^{n}\dfrac{1}{(n+i)^2}\right)=\int_0^1\frac{dx}{(1+x)^2}=-\frac{1}{1+x}\Big|_0^1=-\frac{1}{2}+1=\frac{1}{2}.$$

share|improve this answer
    
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 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 $$\frac{1}{2 n}-\frac{3}{8 n^2}+\frac{7}{48 n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$

share|improve this answer

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

share|improve this answer
    
Uh? How to manage what "error of approximation"? –  DonAntonio Jun 16 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 at 3:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.