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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.

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  • $\begingroup$ 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. $\endgroup$ – Ryan Dec 22 '15 at 3:35
  • $\begingroup$ You've written two different equations above by the way. $\endgroup$ – Elliot G Dec 22 '15 at 3:38
  • $\begingroup$ @ElliotG you're right, sorry! Fixed. $\endgroup$ – feralin Dec 22 '15 at 3:39
  • $\begingroup$ Using the method at this MSE link introduce $$f(z) = \frac{4\pi}{\sin(\pi z)}\frac{1}{1-4z^2}$$ and observe that $$\mathrm{Res}_{z=\pm 1/2} f(z) = -\pi \quad\text{and}\quad \mathrm{Res}_{z=0} f(z) = 4.$$ $\endgroup$ – Marko Riedel Dec 22 '15 at 3:58
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$$\frac{4}{1-4 n^2} = \frac{2}{1-2 n} + \frac{2}{1+2 n}$$

Thus, $$\begin{align}\sum_{n=1}^{\infty} \frac{4 (-1)^n}{1-4 n^2} &= 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2 n-1}+2 \sum_{n=1}^{\infty} \frac{(-1)^n}{2 n+1} \\ &= 2 \frac{\pi}{4} + 2 \left (\frac{\pi}{4}-1 \right ) \\ &= \pi-2 \end{align}$$

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  • $\begingroup$ It always comes back to arctan... $\endgroup$ – Elliot G Dec 22 '15 at 3:39
  • $\begingroup$ How do you get from the right side of the second line to the third line? Is that related to the arc tangent, as @ElliotG mentioned? $\endgroup$ – feralin Dec 22 '15 at 3:41
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    $\begingroup$ @feralin: yes, it follows from a Maclurin expansion of the arctangent. It is also known as the Gregory series for Pi and has been in existence since the 18th century. $\endgroup$ – Ron Gordon Dec 22 '15 at 3:42
  • $\begingroup$ Cool, thanks! I learn something every day... $\endgroup$ – feralin Dec 22 '15 at 3:43
  • $\begingroup$ Interesting is the fact that $$\sum_{n=1}^{\infty} \frac{4(-1)^n}{1-4n^2}x^n=\frac{2 (x+1) \tan ^{-1}\left(\sqrt{x}\right)}{\sqrt{x}}-2$$ $\endgroup$ – Claude Leibovici Dec 22 '15 at 3:54

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