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Suppose i know the following two formulas:

$\displaystyle \sin(\pi z) = \pi z \prod_{n=1}^\infty \left( 1 - \frac{z^2}{n^2} \right)$ and $\pi \cot(\pi z)= \sum_{n=-\infty}^{n=\infty} \frac {1}{z+n}$ Then i am trying to prove that

$\frac {\pi}{\sin{\pi z}}=\sum_{n=-\infty}^{n=\infty} \frac {(-1)^n}{z-n}$

I am not getting the idea for this,any hints/ideas?

  • 0
    The proof is almost the same [with $1/\sin^2(z)$](http://math.stackexchange.com/questions/1959285/sum-k-infty-infty-frac1k-alpha2-frac-pi2-sin2-pi-alpha/1960134#1960134), using the Liouville theorem for showing the function minus its poles is constant. Then adapt it to $\pi / \sin( \pi z)$, being careful with the convergence of $\sum_n \frac{(-1)^n}{z-n}$ – 2017-01-01

1 Answers 1

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Hint. Use Hadamard Factorization Theorem.

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    I think OP want to deduce using the mentioned product and series. – 2017-01-01
  • 0
    The Hadamard factorization theorem follows from this kind of derivation. – 2017-01-01