I need to prove the equation below:
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{4n^{2}-1}=\frac{\pi -2}{4}$$
I think that I can use the following result provided by Fourier Series:
$$|\sin\theta|=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\cos(2n\theta)}{4n^{2}-1} $$
Can I take some value of $\theta$ that will give me the formula? Because $$\cos(2n\theta)=(-1)^{n+1}\Rightarrow 2n\theta=(n+1)\pi\Rightarrow \theta=\frac{\pi}{2}\left(\frac{1}{n}+1\right)$$ does not help me. What can I do?
On
You can use the following, which is a result of Cosine Fourier Series, which is way easier (for your application) than using the Fourier series. The following is valid for $x \in [0,1]$:
\begin{equation} \sin \pi x\simeq \frac{2}{\pi} - \frac{4}{\pi} \sum\limits_{k=1}^{\infty} \frac{1}{4k^2 - 1} \cos( 2 \pi k x) \end{equation} Use $x = \frac{1}{2}$, we get \begin{equation} 1 \simeq \frac{2}{\pi} - \frac{4}{\pi} \overbrace{\sum\limits_{k=1}^{\infty} \frac{1}{4k^2 - 1} (-1)^k}^{\alpha} \end{equation} or \begin{equation} \pi = 2 - 4\alpha \end{equation} so we get $\alpha =\sum\limits_{k=1}^{\infty} \frac{1}{4k^2 - 1} (-1)^k= -\frac{\pi - 2}{4}$. Your summation is $\sum\limits_{k=1}^{\infty} \frac{1}{4k^2 - 1} (-1)^{k+1} = -\alpha.$
You're not taking advantage of periodicity. E.g. your implication should be $2n\theta+2k\pi = (n+1)\pi$, for any integer $k$. Setting $k=n$, gives you $\theta=\pi/2$, so that $\cos(2n\theta)=\cos(n\pi)=(-1)^n$. Can you finish he problem from here?