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In a problem about calculating the Laurent series of an expresion, I'm seeing:

$$\frac{1}{1+\frac{1}{w}}=\sum_{k=0}^\infty \frac{(-1)^k}{w^k}$$ When does that hold? What are the conditions of $1/w$ for that to be true? I guess that in a more general way: $$\sum_{k=0}^\infty (-1)^kz^k=\frac{1}{1+z}$$ Again, what are the conditions on $z$?

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These are geometric series. Do you not know the criterion for convergence of geometric series? Are you familiar with the ratio test? – Gerry Myerson Feb 4 '13 at 11:37
@GerryMyerson Yes, I just saw it, thanks. – MyUserIsThis Feb 4 '13 at 11:38
Then post it as an answer to your question. That is encouraged here. After some time, you can accept it. – Gerry Myerson Feb 4 '13 at 12:26
@GerryMyerson Did it – MyUserIsThis Feb 4 '13 at 14:31

You must have $|z| < 1$. There is a famous joke about the case $z = 1$, there you have $$ 1 - 1 + 1 - 1 + 1 - 1 + ... = \frac{1}{2}$$.

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up vote 1 down vote accepted

Ok, I saw it... It's a geometric serie $$\sum_0^\infty a^k=\frac{1}{1-a},|a|<1$$ In this case, $a=\frac{-1}{z}$, with $|z|>1$, so: $$\frac{1}{1+\frac{1}{z}}=\frac{1}{1-\frac{-1}{z}}=\sum_{k=0}^\infty \left( \frac{-1}{z}\right)^k=\sum_{k=0}^\infty \frac{(-1)^k}{z^k}$$

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