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Find all z such that the series (-1)n / (z+n) converges

by RoRi

Find all complex numbers z such that the series

    \[ \sum_{n=1}^{\infty} \frac{(-1)^n}{z+n} \]

converges.

First, the series is not absolutely convergent for any z \in \mathbb{Z} since

    \[ \sum_{n=1}^{\infty} \left| \frac{(-1)^n}{z+n} \right| = \sum_{n=1}^{\infty} \frac{1}{|z+n|}. \]

By limit comparison with \sum \frac{1}{n} this always diverges.

Then, if z \neq -n for any positive integer n the series is convergent since \frac{1}{z+n} is monotonically decreasing and \lim_{n \to \infty} \frac{1}{z+n} = 0. If z = -n for some positive integer n, then the series is not defined since it is undefined for that term.

2 comments

  1. Anonymous says:

    Let 

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    , and the real and imaginary parts of the original series are alternating series(es), which converge by Lebniz’s rule.

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