Show identity using fourier-coefficents.

Question

I have to show that $$\sum_{i=1}^{\infty}\frac{(-1)^n}{4n^2-1} = \frac{2-\pi}{4}$$ using the fourier-coefficents of $\cos(\frac{x}{2}), x \in ]-\pi,\pi[$. I know that $c_0 = \frac{2}{\pi}$ and that $$c_n = \frac{(-1)^n}{2\pi(\frac14-n^2)}$$ and I can't get it right. Can somebody help?

Answer 1

For a check of the result:

Note the Gregory series: $\tan^{-1}1=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}...=\frac{\pi}{4}.$

Use $\frac{(-1)^n}{4n^2-1}=\frac{(-1)^n}{2}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]$.

Then $\sum_{i=1}^{\infty}\frac{(-1)^n}{4n^2-1}=-\frac{1}{2}+[\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}...]=\frac{2-\pi}{4}.$

Answer 2

Set $f(x)=\cos\left(\frac{x}{2}\right)$, and notice that $f$ is a smooth function, and so its Fourier series converges to it. Furthermore $f$ is an even function, so we can write it as a cosine series, and in particular, as $f$ has cosine Fourier coefficients (you seem to have gotten these slightly wrong)

$$c_0=\frac{4}{\pi}, \quad c_n=-\frac{4}{\pi}\frac{(-1)^n}{4n^2-1},$$

we have that

$$\cos\left(\frac{x}{2}\right)=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n=1}^\infty\frac{(-1)^n}{4n^2-1}\cos(nx).$$

Now setting $x=0$ this yields the equality

$$1=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n=1}^\infty\frac{(-1)^n}{4n^2-1}.$$

Rearranging we get that

$$\sum_{n=1}^\infty\frac{(-1)^n}{4n^2-1}=\frac{2-\pi}{4},$$

which is the desired result.