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

I recently found a series representation for 1 from the calculation of a Fourier series: $$1 = \frac{2}{\pi} + \sum_{n=1}^{\infty} \frac{4(-1)^n}{\pi(1-4n^2)}$$ From this, I can easily find that $$\sum_{n=1}^{\infty} \frac{4(-1)^n}{1-4n^2} = \pi - 2,$$ but what other methods are there to evaluate the sum? I don't even know where to start.

• It looks like the denominator can be split by factoring and using partial fractions. Trying different values for $n$ I can see a definite pattern. – Ryan Dec 22 '15 at 3:35
• You've written two different equations above by the way. – Elliot G Dec 22 '15 at 3:38
• @ElliotG you're right, sorry! Fixed. – feralin Dec 22 '15 at 3:39
• Using the method at this MSE link introduce $$f(z) = \frac{4\pi}{\sin(\pi z)}\frac{1}{1-4z^2}$$ and observe that $$\mathrm{Res}_{z=\pm 1/2} f(z) = -\pi \quad\text{and}\quad \mathrm{Res}_{z=0} f(z) = 4.$$ – Marko Riedel Dec 22 '15 at 3:58

$$\frac{4}{1-4 n^2} = \frac{2}{1-2 n} + \frac{2}{1+2 n}$$
Thus, \begin{align}\sum_{n=1}^{\infty} \frac{4 (-1)^n}{1-4 n^2} &= 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2 n-1}+2 \sum_{n=1}^{\infty} \frac{(-1)^n}{2 n+1} \\ &= 2 \frac{\pi}{4} + 2 \left (\frac{\pi}{4}-1 \right ) \\ &= \pi-2 \end{align}
• Interesting is the fact that $$\sum_{n=1}^{\infty} \frac{4(-1)^n}{1-4n^2}x^n=\frac{2 (x+1) \tan ^{-1}\left(\sqrt{x}\right)}{\sqrt{x}}-2$$ – Claude Leibovici Dec 22 '15 at 3:54