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Given $\sum \limits_{k\ge 1} \frac{(-1)^{k-1}}{z+k}$, how do we find the function f that corresponds to this series? How do we find the taylor series of f in the point $z=0$ ?

Thank you very much for every input. xD

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up vote 3 down vote accepted

Let's split the sum into even and odd parts: $\sum \limits_{k \geqslant 1} \frac{(-1)^{k-1}}{z+k} = \sum \limits_{k \geqslant 0} \left( \frac{1}{z+2 k +1} - \frac{1}{z+2 k +2} \right) = \sum_{k \geqslant 0} \left( \left( \frac{1}{z+2 k +1} - \frac{1}{2k+2} \right) - \left( \frac{1}{z+2 k +2} - \frac{1}{2k+2} \right) \right)$.

Notice that $\sum \limits_{k \geqslant 0} \left( \frac{1}{z+2 k +1} - \frac{1}{2k+2} \right) = \frac{1}{2} \sum \limits_{k \geqslant 0} \left( \frac{1}{k + (1+z)/2} - \frac{1}{k+1} \right) = -\frac{1}{2} \psi\left(\frac{1+z}{2} \right)$, where $\psi(z)$ is the digamma function.

Likewise $\sum \limits_{k \geqslant 0} \left( \frac{1}{z+2 k +2} - \frac{1}{2k+2} \right) = -\frac{1}{2} \psi\left(\frac{z}{2} + 1\right) $.

Thus $$ \sum_{k \geqslant 1} \frac{(-1)^{k-1}}{z+k} = \frac{1}{2} \left( \psi\left(\frac{z}{2} + 1\right) - \psi\left(\frac{1+z}{2} \right) \right) $$

The Taylor series around $z=0$ actually follows from the defining series: $$ [z]^m \sum_{k \geqslant 1} \frac{(-1)^{k-1}}{z+k} = \sum_{k \geqslant 1} \frac{(-1)^{k+m+1}}{k^m} = (-1)^{m+1} \left(1- 2^{-m} \right) \zeta(m+1) $$ thus $$ \frac{1}{2} \left( \psi\left(\frac{z}{2} + 1\right) - \psi\left(\frac{1+z}{2} \right) \right) = \sum_{m=0}^\infty (-1)^{m+1} \left(1- 2^{-m} \right) \zeta(m+1) z^m $$

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It's a shame that the digamma function is notated with a psi. – alex.jordan Dec 6 '11 at 22:25

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