Calculate the Laurent series for $\displaystyle\frac{z}{z+1}$ when $|z|>1$.
There is really no singularity here, right? Can I just use a Taylor series, or what should I do?
2026-04-01 17:49:19.1775065759
Laurent series for $\frac{z}{z+1}$ on $|z|>1$
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For $|z|\gt1$, $$\frac{z}{z+1}=\frac1{1+\frac1z}=\sum_{n\geqslant0}\frac{(-1)^n}{z^n}.$$ Edit: For $|z|\lt1$, $$\frac{z}{z+1}=\sum_{n\geqslant0}(-1)^nz^{n+1}.$$