I must show that $\sum\limits_{n\geq1} \frac{\sin(nx)}{n^2}$ converges $\forall x \in {R}$. Then if $f(x)=\sum\limits_{n=1}^\infty f_n(x)$, I must prove that $f(x)$ is continuous for $x\in [0, \pi]$ and $\int\limits_0^\pi f(x)=2\sum\limits_{n=1}^\infty \frac{1}{(2n-1)^3}$.

I used Weistrass M-test: $\sum\limits_{n\geq1} \frac{\sin(nx)}{n^2}$ converges uniformly because $|\frac{\sin(nx)}{n^2}|\le \frac{1}{n^2}, \forall x\in R$ and $\sum \frac {1}{n^2}$ converges.

I found somewhere that $f(x)=\lim\limits_{n\to\infty}f_n(x)$. Where does it come from? I don't understand. How $f(x)$ looks like? Without $f(x)$ I don't know how to prove the last equality from exercise:(

Later edit: If $ \sum\limits_{n\geq1}f_n\to f$ uniformly , then $f_n\to f$ (punctual converges to $f $- I hope this is correct translation) which means $f_n(x)\to f(x)$ when $n\to \infty ?$

I hope someone could help me again...

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    $\begingroup$ Convergence $\sum_{n\ge 1} f_n \to f$ can not imply $f_n\to f$, otherwise the series would be divergent. $\endgroup$ – Sasha Aug 11 '11 at 14:54
  • $\begingroup$ This is a Clausen function: en.wikipedia.org/wiki/Clausen_function $\endgroup$ – graveolensa Aug 11 '11 at 18:25
  • $\begingroup$ "converges pointwise to f" is the standard translation, but "punctual convergence" is perfectly comprehensible as the same thing. $\endgroup$ – zyx Aug 22 '11 at 6:15

As you observed, by the Weierstrass M-test, the series converges uniformly. Let $$f(x)=\sum_1^\infty \frac{\sin(nx)}{n^2}.$$ The functions $\sin(nx)/n^2$ are continuous, so their "sum" $f(x)$ is continuous. (In general, if we do not have uniform convergence, but only pointwise convergence, the sum need not be continuous.)

The above result about continuity is causing you some confusion. It is based on a theorem that says, or should say, that if $(f_n(x))$ is a sequence of continuous functions that converges uniformly to $f(x)$ in an interval, then $f(x)$ is continuous.

To apply the theorem to a series $\sum_{n \ge 1} g_n(x)$, we just let $f_n(x)=\sum_{k=1}^n g_k(x)$. There is nothing new in this: convergence of infinite series is defined in terms of what happens to partial sums as $n$ goes to infinity.

The additional fact that is needed is that a series uniformly convergent on a finite interval can be integrated term by term on that interval. So we need to calculate $\int_0^\pi \frac{\sin(nx)}{n^2} dx$ and "add up."

More explicitly, $$\int_0^\pi \left(\sum_{n\ge 1} \frac{\sin(nx)}{n^2}\right)\,dx=\sum_{n \ge 1} \left(\int_0^\pi \frac{\sin(nx)}{n^2}\,dx\right).$$ (The above interchange of the order of integration and summation is permitted because of the uniform convergence. It can fail when we do not have uniform convergence.)

The integration is straightforward, for $\sin(nx)$ has $-\dfrac{\cos(nx)}{n}$ as an antiderivative.
Thus $$\int_0^\pi \frac{\sin nx}{n^2}dx=\frac{1}{n^3}(-\cos(n\pi)+1).$$ If $n$ is even, $\cos(n\pi)=1$, so the integral is $0$, and contributes nothing to the ultimate sum.

If $n$ is odd, then $\cos(n\pi)=-1$, so the integral is $\dfrac{2}{n^3}$. Since $n$ is odd, let $n=2m-1$. Then $$\int_0^\pi \frac{\sin nx}{n^2}dx=\frac{2}{(2m-1)^3}.$$ We conclude that $$\int_0^\pi f(x) \;dx=\sum_1^\infty \frac{2}{(2m-1)^3}.$$ (In your post, the final variable of summation is $n$, but that makes no difference.)

Interesting fact: The sum that we end up with is closely related to $\sum_{n \ge 1} \frac{1}{n^3}$. This last sum, also known as $\zeta(3)$, was proved to be irrational by Apery about $30$ years ago, using an elementary (but not easy) argument. It was one of those rare mathematical results that even gets reported in the mainstream press.

  • $\begingroup$ Thank you very much André Nicolas. The first part of your explanations helps me a lot. $\endgroup$ – NumLock Aug 11 '11 at 15:18
  • $\begingroup$ In the last part of your explanation you wrote cos(nπ) instead of sin(nπ) :P Please change that. Thank you again for your help! $\endgroup$ – NumLock Aug 11 '11 at 15:59
  • $\begingroup$ @Numlock: Thanks for spotting the inconsistency. It really is $\cos n\pi)$. But the function I am integrating is $\sin(nx)$, and in a couple of places it was written as $\cos(nx)$. Fixed! Someday I will write an answer free of things like that. $\endgroup$ – André Nicolas Aug 11 '11 at 16:08
  • $\begingroup$ Perhaps for the sake of completeness we could include the the interchange of limits is justified by uniform convergence since we are integrating over a finite interval. Over infinite integrals, uniform convergence alone does not let us switch the limits. I have not been able to forget this detail since someone pointed out that my solution to a previous Putnam problem was incorrect because of this! $\endgroup$ – Ragib Zaman Aug 22 '11 at 3:37
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    $\begingroup$ @Ragib Zaman: Thank you for the suggested change, I will try to change wording so that no one is led astray on a Putnam! $\endgroup$ – André Nicolas Aug 22 '11 at 5:17

Differentiate twice, get the series for $-\dfrac{1}{2 \tan\frac{x}{2}}$. So now integrate this twice... Of course that is not elementary, but still...


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