This is the problem 1.3.8 from Powers' Boundary Value Problems, 5th Edition. It says:
The series $$ \sum_{n=1}^\infty \frac{1}{n^3} \, \sin(nx) $$ converges to a continuous periodic function. On the interval $0 < x < 2\pi$, this function coincides with a polynomial $p(x)$ of degree $3$. Find the polynomial.
This is my attempt so far:
I started setting the polynomial as $$p(x)=A+Bx+Cx^2+Dx^3.$$ Since the sum only contains $\sin$ factors, it must be an odd function, so it reduces to $$p(x)= Bx+Dx^3$$ Then (as the problem suggest) I look for the points where $p(x)=0$, which are $x=0$ and $x=\pm\sqrt{-\frac{B}{D}}$. Then, evaluating this last point in the series: $$ 0 = p\left(\pm\sqrt{-\frac{B}{D}}\right) = \sum_{n=1}^\infty \dfrac{1}{n^3} \, \sin\left(n\left(\pm\sqrt{-\frac{B}{D}}\right)\right) $$ which implies that $$ \left(\pm\sqrt{-\frac{B}{D}}\right) = \pi\Longrightarrow B = -D\pi^2 $$ so now the function has the form $$ p(x) = (-D\pi^2)x+Dx^3 $$ Then, evaluating in the point $x=\frac{\pi}{2}$, it turns out that $D=-\frac{1}{12}$, so $$ p(x) = \left(\frac{\pi^2}{12}\right)x - \left(\frac{1}{12}\right)x^3 $$
This is all I've tried, but I'm not sure about the results... am I missing something? Is there anything I'm taking for granted?
Any help will be appreciated. Thank you in advance.