# Series representation for 1/cos x

Why does the summation

\begin{equation*} \frac{1}{\cos x}=\sum_{n=1}^\infty \frac{(-1)^n(2n-1)\pi }{x^2-\left (n-\frac{1}{2}\right )^2\pi^2} \end{equation*}

hold?

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You can find such partial fraction expansions given as examples in many texts on complex analysis. For example: books.google.com/…. It uses "Cauchy's theorem on partial fraction expansions": books.google.com/…. The analogous series for $\tan$ is given in Wikipedia: en.wikipedia.org/wiki/… – Jonas Meyer Oct 29 '10 at 22:05
P.S.: I'd upvote your question, but I used up all my votes for the day. However, it might help to add a bit about the context in which you came across this formula and what level of mathematics you'd like in an answer. – Jonas Meyer Oct 29 '10 at 22:14
I found it unsolved in AOPS and I was just interested how to prove that. Personally, in mathematics I like very rigorous proofs but if no such proof is found, I'm usually grateful for sketches too. – Jaska Oct 29 '10 at 22:21
Thank you Jaska. There is a rigorous proof given in the book I linked above, which I just left as a comment because I am not currently inclined to say more. By "level", I meant, for example, are you comfortable with the theory of complex analytic functions? – Jonas Meyer Oct 29 '10 at 22:26
Yes. I have read the basic course of complex analysis in the university of Helsinki so I know what are complex analytic functions. – Jaska Oct 29 '10 at 22:31

Example 2 gives the series for $\cot$, although from the preview I can't see whether $\tan$ and $\csc$ are included. The analogous series for $\tan$ is derived on Wikipedia. Each of these series can be useful for evaluating numerical series. For example, taking $x=0$ in your series you get $$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots.$$ And although it is not directly related to your question, I thought I'd add in light of the recent question on evaluating the sum of the reciprocals of the squares that if you take the partial fraction series $$\tan(x) = \sum_{k=0}^{\infty} \frac{-2x}{x^2 - \left(k + \frac{1}{2}\right)^2\pi^2},$$ divide by $x$ and let $x$ go to $0$, rearranging yields $$\frac{\pi^2}{8}=1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+\cdots,$$ which in turn (because the sum over the evens is $\frac{1}{4}$ the total sum) leads to $$\frac{\pi^2}{6}=1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\cdots.$$