<s>[INST] How to use mittag-leffer expansion to compute $\texttt{sech} z$? A question concerning a paper from Kesten (1986) In Kesten's 1986 paper (Limit distribution of Sinai's Random Walk) we read:

The reference  is E.T. Whittaker and G.N. Watson, A Course of Modern Analysis, 4th ed. (Cambridge Univ. Press, Cambridge, 1952). 
We see that:

Using this result we can begin to do some manipulations:
First write
 $$\frac{1}{\cosh(\sqrt{2\theta})} = \texttt{sech} \sqrt{2\theta} = \sum_{k=0}^\infty \frac{4\pi (2k + 1) (-1)^k}{(2k+1)^2\pi^2 + 8\theta} $$
Then we come back to the expression we are interested in:
$$\Bbb{E}\{ e^{-\theta L}; L>0\} =\frac{1}{2\theta} \bigg[1 - \texttt{sech}(\sqrt{2\theta})\bigg]\\
= \frac{1}{2\theta} \bigg[1 - \sum_{k=0}^\infty \frac{4\pi (2k + 1) (-1)^k}{(2k+1)^2\pi^2 + 8\theta}\bigg]  $$
But how do we obtain the second line
$$ \frac{16}{\pi} \sum_{k=0}^\infty    \frac{(-1)^k}{(2k + 1)}\frac{1}{(2k+1)^2\pi^2 + 8\theta} ?$$
 [/INST] Note first that $\texttt{cosh} (0) = 1$ so
$$ 1 = \frac{1}{\cosh(0)} = \texttt{sech} (0) = \sum_{k=0}^\infty \frac{4\pi (2k + 1) (-1)^k}{(2k+1)^2\pi^2 }$$
Therefore
$$ \Bbb{E}\{ e^{-\theta L}; L>0\} =\frac{1}{2\theta} \bigg[1 - \texttt{sech}(\sqrt{2\theta})\bigg]\\
= \frac{1}{2\theta} \bigg[ \sum_{k=0}^\infty \frac{4\pi (2k + 1) (-1)^k}{(2k+1)^2\pi^2 } - \sum_{k=0}^\infty \frac{4\pi (2k + 1) (-1)^k}{(2k+1)^2\pi^2 + 8\theta}\bigg]\\
=\frac{4\pi}{2\theta}\bigg[\sum_{k=0}^\infty \frac{8\theta (2k + 1) (-1)^k }{(2k+1)^2\pi^2} \frac{1}{(2k+1)^2\pi^2 + 8\theta}\bigg]\\
=\frac{16}{\pi}\bigg[\sum_{k=0}^\infty \frac{ (-1)^k }{(2k+1)} \frac{1}{(2k+1)^2\pi^2 + 8\theta}\bigg]$$
which is the desired result
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