Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Find $$ \lim\limits_{n\to\infty}\sum\limits_{i=1}^{n}{\frac{2n}{(n+2i)^2}}.$$

I have tried dividing through by $1/n^2$ and various other algebraic tricks but cannot seem to make any progress on this limit. Wolfram Alpha gives the value as $2/3$, but I could not understand its derivation. Any insight is welcome.

share|cite|improve this question
up vote 20 down vote accepted

The expression $$ \sum_{i = 1}^n \frac{2n}{(n+2i)^2} =\frac{1}{n} \sum_{i = 1}^n \frac{2}{(1+\frac{2i}{n})^2} $$ is the Riemann sum of the function $f(x)= \frac{2}{(1+2x)^2}$ over the interval $I = [0,1]$, corresponding to the uniform partition of $I$ into $n$ equal parts. Since $f$ is Riemann-integrable (being a continuous function over a closed and bounded interval), this sum approaches the integral of $f$ over $I$ as $n \to \infty$. That is, $$ \begin{eqnarray*} \lim_{n \to \infty} \frac{1}{n} \sum_{i = 1}^n \frac{2}{(1+\frac{2i}{n})^2} &=& \int_0^1 \frac{2}{(1+2x)^2} \mathrm{d} \, x \\ &=& \left. -\frac{1}{1+2x} \right|_{x=0}^{x=1} \quad = \quad \frac{2}{3}. \end{eqnarray*} $$

share|cite|improve this answer
+1: I really like this approach. – Mike Spivey Oct 29 '11 at 5:54
Thanks @Mike. ${}{}$ – Srivatsan Oct 29 '11 at 5:54
Thanks @Daniel, for coming back to accept an answer. Many users forget to do that... =) – Srivatsan Oct 31 '11 at 14:45

You can also use Euler-Maclaurin summation.

The first-order Euler-Maclaurin formula says $$\sum_{i=1}^n f(i) = \int_1^n f(x) \, dx + {f(1) + f(n) \over 2} + \int_1^n f'(x) \left(x - \lfloor x \rfloor - \frac{1}{2}\right)\,dx.$$

Since $|x - \lfloor x \rfloor - \frac{1}{2}| \leq 1$, with $f(x) = \frac{2n}{(n+2x)^2}$ we have $$\sum_{i=1}^n \frac{2n}{(n+2i)^2} = \int_1^n \frac{2n \, dx}{(n+2x)^2} + R_n,$$ where $|R_n| \leq \left|\frac{3f(n) - f(1)}{2}\right| = \left|\frac{3}{n} - \frac{n}{(n+2)^2}\right|$.

Therefore, $$\lim_{n \to \infty} \sum_{i=1}^n \frac{2n}{(n+2i)^2} = \lim_{n \to \infty} \int_1^n \frac{2n \, dx}{(n+2x)^2} = \lim_{n \to \infty} \left[\frac{-n }{n+2x} \right]_1^n = \lim_{n \to \infty} \left(-\frac{1}{3} + \frac{n}{n+2}\right) = \frac{2}{3}.$$

share|cite|improve this answer
+1 Your solution is also nice. I especially like the fact that this yields an error bound of $O(1/n)$ as well. – Srivatsan Oct 29 '11 at 5:58
This solution is good, but we need the best one which is simpler than this one. – Hassan Muhammad Oct 29 '11 at 6:19
@Hassan: I am missing the point of your comment... – J. M. Oct 29 '11 at 15:37

Your Answer


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.