I am learning Fourier Series and I cannot figure this question out,
Find the value of the infinite sum,
$1 + \frac{1}{2} – \frac{1}{4} – \frac{1}{5} + \frac{1}{7} + \frac{1}{8} – \frac{1}{10} – \frac{1}{11} + …
$
I am guessing that you have to say that the sum of the series is $f(x)$ and that
$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n cos(\frac{n\pi x}{l}) b_n sin(\frac{n\pi x}{l})$
and that
$a_n = \frac{1}{l} \int_{-l}^{l} f(x) cos(\frac{n\pi x}{l})dx$
$b_n = \frac{1}{l} \int_{-l}^{l} f(x) sin(\frac{n\pi x}{l})dx$
So then I tried to say that
$\frac{1}{2} = a_1 = \frac{1}{l} \int_{-l}^{l} f(x) cos(\frac{\pi x}{l})dx * cos(\frac{\pi x}{l})$
But I am not too sure here or even sure that I am going in the right direction. Does anyone know how to proceed with this problem?
Best Answer
This is not exact science, there is a bit of experimental mathematics here.
As I pointed out in my comment, the shape of the terms suggests working with the sum $$S(x)=\sum_{n=1}^\infty \frac{\sin nx}{x}$$ and take the value at $\frac{\pi}{3}$.
Problem is : what is the function $S$ ? Doing a little research with Wolfram Alpha and Geogebra, you can infer that the function $$f:x\mapsto \begin{cases}\frac{\pi-x}{2}&\text{ if }x\in]0,\pi] \\ \frac{x+\pi}{2}&\text{ if }x\in[-\pi,0[\end{cases}$$ prolonged by $2\pi$-periodicity could do the trick.
$f$ is continuously differentiable by parts, so its Fourier series converges at every continuity point towards $f$. The coefficients are : $$b_n=\frac{2}{\pi}\int_0^\pi \frac{x+\pi}{2}\sin nx{\rm d}x = \frac{1}{n}$$ so $f(x)=S(x)$ except at points of the form $2k\pi$.
Taking the value at $\frac{\pi}{3}$, you find : $S(\pi/3)=\frac{\pi}{3}$, but also $$S(3)=\sum_{n=1}^\infty \frac{\sin n\frac\pi3}{n} = \frac{\sqrt3}{2}(1+\frac12-\frac14-\frac15+\frac17+\frac18-\dots)$$ so your sum evaluates to $\frac2{\sqrt3}\frac\pi3=\frac{2\pi}{3\sqrt3}$, as mentionned by Brevan.