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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to evaluate this limit?

$$\underset{n\to \infty }{\mathop{\lim }}\,\left( {{n}^{-2}}\sum\limits_{i=1}^{n}{\sum\limits_{j=1}^{{{n}^{2}}}{\frac{1}{\sqrt{{{n}^{2}}+ni+j}}}} \right)$$

Thanks!

share|cite|improve this question
up vote 3 down vote accepted

The idea here is to rearrange in terms of a Riemann sum that leads to a double integral. Here, a little manipulation produces

$$\frac{1}{n} \sum_{i=1}^n \frac{1}{n^2} \sum_{j=1}^{n^2} \frac{1}{\sqrt{1+\frac{i}{n}+\frac{j}{n^2}}}$$

Then as $n \to \infty$, the double sum becomes

$$\begin{align}\int_0^1 dx \, \int_0^1 dy \frac{1}{\sqrt{1+x+y}} &= 2 \int_0^1 dx \left [ \sqrt{2+x}-\sqrt{1+x}\right ]\\ &= \frac{4}{3} \left [\left (3^{3/2}-2^{3/2} \right ) - \left (2^{3/2}-1 \right )\right ]\\&=4 \sqrt{3} - \frac{16}{3} \sqrt{2}+1\end{align}$$

share|cite|improve this answer
    
Thank you very much! – Sleepingip May 27 '13 at 12:59

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.