Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question
add comment

1 Answer

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|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
add comment

Your Answer

 
discard

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.