Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given $$\frac{\pi}{\sin (\pi z)}=\sum_{n=-\infty}^\infty (-1)^n \frac{1}{z-n},$$ is there a fast way to get $$\sec(z)=\sum_{n=1}^\infty \frac{(-1)^n(2n-1)\pi}{z^2-(n-1/2)^2\pi^2}?$$ I've tried computing it, but I get a huge mess. Would anybody have some ideas? Thanks in advance.

share|cite|improve this question
up vote 2 down vote accepted

Let's rewrite your first equation as : $$\frac{\pi}{\cos(\pi/2-\pi z)}=\sum_{n=-\infty}^\infty \frac{(-1)^n}{z-n}$$ and set $\ x:=\pi\bigl(\frac 12-z\bigr)\ $ then (dividing by $\pi$) : $$\sec(x)=\sum_{n=-\infty}^\infty \frac {(-1)^n}{\pi z-\pi/2-(n-1/2)\pi}$$ for $m:=-n$ : $$\sec(x)=-\sum_{n=1}^\infty \frac {(-1)^n}{x+(n-1/2)\pi}-\sum_{m=0}^\infty \frac {(-1)^m}{x+(-m-1/2)\pi}$$ for $n':=m+1$ : $$\sec(x)=-\sum_{n=1}^\infty \frac {(-1)^n}{x+(n-1/2)\pi}+\sum_{n'=1}^\infty \frac {(-1)^{n'}}{x-(n'-1/2)\pi}$$ Putting everything together we conclude : $$\sec(x)=\sum_{n=1}^\infty \frac{(-1)^n(2n-1)\pi}{x^2-(n-1/2)^2\pi^2}.$$

share|cite|improve this answer
Perfect! Thanks a lot, this is much better than my reckless bashing. – mathstudent12 Oct 11 '12 at 0:18
@mathstudent12: Thanks for that! The idea here was to put the positive half-integers at the left and the negative ones at the right. – Raymond Manzoni Oct 11 '12 at 19:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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