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.

  • $\begingroup$ 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

$$\frac{4}{1-4 n^2} = \frac{2}{1-2 n} + \frac{2}{1+2 n}$$

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.