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.


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, $$ \begin{eqnarray*} \lim_{n \to \infty} \frac{1}{n} \sum_{i = 1}^n \frac{2}{(1+\frac{2i}{n})^2} &=& \int_0^1 \frac{2}{(1+2x)^2} \mathrm{d} \, x \\ &=& \left. -\frac{1}{1+2x} \right|_{x=0}^{x=1} \quad = \quad \frac{2}{3}. \end{eqnarray*} $$

  • 1
    $\begingroup$ Thanks @Daniel, for coming back to accept an answer. Many users forget to do that... =) $\endgroup$ – 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, $$\lim_{n \to \infty} \sum_{i=1}^n \frac{2n}{(n+2i)^2} = \lim_{n \to \infty} \int_1^n \frac{2n \, dx}{(n+2x)^2} = \lim_{n \to \infty} \left[\frac{-n }{n+2x} \right]_1^n = \lim_{n \to \infty} \left(-\frac{1}{3} + \frac{n}{n+2}\right) = \frac{2}{3}.$$

  • 3
    $\begingroup$ +1 Your solution is also nice. I especially like the fact that this yields an error bound of $O(1/n)$ as well. $\endgroup$ – Srivatsan Oct 29 '11 at 5:58
  • $\begingroup$ This solution is good, but we need the best one which is simpler than this one. $\endgroup$ – Hassan Muhammad Oct 29 '11 at 6:19
  • 7
    $\begingroup$ @Hassan: I am missing the point of your comment... $\endgroup$ – J. M. is a poor mathematician Oct 29 '11 at 15:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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