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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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 03:35
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You've written two different equations above by the way. – pancini Dec 22 '15 at 03:38
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@ElliotG you're right, sorry! Fixed. – feralin Dec 22 '15 at 03:39
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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 03:58
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$$\frac{4}{1-4 n^2} = \frac{2}{1-2 n} + \frac{2}{1+2 n}$$
Thus, $$\begin{align}\sum_{n=1}^{\infty} \frac{4 (-1)^n}{1-4 n^2} &= 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2 n-1}+2 \sum_{n=1}^{\infty} \frac{(-1)^n}{2 n+1} \\ &= 2 \frac{\pi}{4} + 2 \left (\frac{\pi}{4}-1 \right ) \\ &= \pi-2 \end{align}$$
Ron Gordon
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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? – feralin Dec 22 '15 at 03:41
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1@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. – Ron Gordon Dec 22 '15 at 03:42
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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 03:54
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@GohP.iHan. No idea ! I just tried with a CAS and got this funny result. – Claude Leibovici Dec 25 '15 at 11:54
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@ClaudeLeibovici well it's true for $ x=1 $ by applying Abel's Theorem. Do you mind if I ask this problem in MSE? – GohP.iHan Dec 25 '15 at 14:17
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@GohP.iHan. Why do you want me to mind ? Just go ahead. By the way, Merry Xmas. Cheers :-) – Claude Leibovici Dec 25 '15 at 14:30
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