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According to the Mittag-Leffler partial fraction expansion theorem for meromorphic functons, $$ \frac{1}{\cos z}= 1+\sum_{n=-\infty}^{\infty}(-1)^{n+1}\Big (\frac{1}{z-\frac{\pi}{2}-\pi n}+\frac{1}{\frac{\pi}{2}+\pi n}\Big). $$

But how does one rearrange terms (and justify the rearrangement) to show that $$\frac{1}{\cos z} = \sum_{n=0}^{\infty}(-1)^{n+1}\frac{(2n+1)\pi}{z^{2}-(\frac{2n+1}{2})^{2}\pi^2} \ ? $$

EDIT:

An alternative version of the theorem (which requires showing that $\int \frac{\sec w}{w-z} \ dw$ vanishes around an appropriate contour) allows one to conclude that

$$ \begin{align} \frac{1}{\cos z} &= \sum_{n=-\infty}^{\infty} \frac{(-1)^{n+1}}{z - \frac{\pi}{2}- \pi n} \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{(-n-1)+1}}{z - \frac{\pi}{2} - \pi (-n-1)} + \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{z - \frac{\pi}{2} - \pi n} \\ &= \sum_{n=0}^{\infty} (-1)^{n+1} \left( \frac{-1}{z+ \frac{\pi}{2} + \pi n} + \frac{1}{z- \frac{\pi}{2} - \pi n} \right) \\ &= \sum_{n=0}^{\infty} (-1)^{n+1} \frac{(2n+1) \pi}{z^{2}- (\frac{2n+1}{2})^{2}\pi^{2}}. \end{align}$$

Random Variable
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2 Answers2

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The terms in $\Big (\frac{1}{z-\frac{\pi}{2}-\pi n}+\frac{1}{\frac{\pi}{2}+\pi n}\Big)$ are not absolutely convergent, since $\sum_{n=1}^{\infty} \frac1{z-\frac{\pi}{2}-\pi n}$ diverges.

So you may have problems.

marty cohen
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Hint: Let $$a_n = (-1)^{n+1}\left(\frac{1}{z-\pi/2-n\pi} + \frac{1}{\pi/2+n\pi}\right).$$ Notice that $a_n = O(1/n^2)\, (n\to\infty)$ and that $$a_n + a_{-n-1} = (-1)^{n+1}\frac{(2n+1)\pi}{z^2-\left(\frac{2n+1}{2}\right)^2\pi^2} + (-1)^{n+1}\frac{4}{(2n+1)\pi}.$$

user26872
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    So $$\sum_{n=-\infty}^{\infty} a_{n} = \sum_{n=0}^{\infty} a_{n} + \sum_{n=0}^{\infty} a_{-n-1} = \sum_{n=0}^{\infty} (a_{n}+a_{-n-1}) = \sum_{n=0}^{\infty}(-1)^{n+1} \Big( \frac{(2n+1)\pi}{z^2-\left(\frac{2n+1}{2}\right)^2\pi^2} + \frac{4}{(2n+1)\pi} \Big) = \sum_{n=0}^{\infty}(-1)^{n+1} \frac{(2n+1)\pi}{z^2-\left(\frac{2n+1}{2}\right)^2\pi^2} - \frac{4}{\pi} \arctan(1) = \sum_{n=0}^{\infty}(-1)^{n+1} \frac{(2n+1)\pi}{z^2-\left(\frac{2n+1}{2}\right)^2\pi^2} -1 ?$$ – Random Variable Jul 27 '13 at 05:17
  • @RandomVariable: Looks good. – user26872 Jul 27 '13 at 05:24
  • What is the significance of the behavior of $a_{n}$ at infinity? – Random Variable Jul 27 '13 at 05:35
  • @RandomVariable: It justifies the rearrangement $\sum_{n=-\infty}^\infty a_n = \sum_{n=0}^\infty (a_n+a_{-n-1})$. At present I can't justify the second rearrangement (leading to the final result). – user26872 Jul 27 '13 at 05:51