Finding a function from a fourier series

Question

Taken from Apostol Analysis, it says, find a continuous function that generates the fourier series:

$$ \sum_{n} \frac{\left(-1\right)^n}{n^3} \sin(nx) $$

I really have no idea how to solve this, instinctively I tried solving $\langle f,\sin(nx)\rangle = \frac{(-1)^n}{n^3}$ and $\langle f,\cos(nx)\rangle = 0$ and got nowhere, any help would be much appreciated.

Answer 1

Observations:

1. $f$ is odd
2. $f$ is continuous on $\mathbb R$, because the series converges uniformly. Together with 1, this implies $f(\pi)=0$.
3. $f$ is a cubic polynomial on $(-\pi,\pi)$, because the Fourier coefficients of $x^k$ involve $1/n^k$ (integration by parts happens $k$ times).

Odd cubic polynomials vanishing at $\pi$ are of the form $A(x^3-\pi^2 x)$. There are various ways to find $A$, including boring integration. A less boring way is to observe that
$$ f'(x)= \sum_{n=1}^\infty \frac{(-1)^{n}}{n^2}\cos nx$$ is continuous on $\mathbb R$ and $$f'(\pi)= \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi ^2}{6}$$ Since $A(x^3-\pi^2 x)'=A(3x^2-\pi^2)$ evaluates to $2\pi^2A$ at $x=\pi$, we have $A=1/12$.