Can somebody help me with determining the radius of convergence of the power series of the following function
$$f(z)=\frac{\mathrm{e}^z}{z-1}$$ about $z=0$?
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5What could (and does) stop the power series from converging for that function? – Daniel Fischer Jan 23 '14 at 20:51
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1The power series about $z=$ what? – Jan 23 '14 at 20:51
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I guess at $z=1$... :D – Jlamprong Jan 23 '14 at 20:52
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Try $z \mapsto {1 \over z-1}$ first. – copper.hat Jan 23 '14 at 20:53
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A related problem. – Mhenni Benghorbal Jan 23 '14 at 21:06
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1RoosJansen, you have your answer already, but in the future it's usually acceptable to specify your knowledge of basic concepts the problem involves and attempts to solve it; beyond the fact that some users might prefer helping you over doing the problem for you, it actually helps members tailor the answer to your needs (putting emphasis on whichever part you're not comfortable with, instead of just coming up with their own 'shortest-path' to the goal). – Jonathan Y. Jan 23 '14 at 21:43
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Note that $$ f(z)=\frac{\mathrm{e}^z}{z-1}=\frac{\mathrm{e}^z-e}{z-1}+\frac{e}{z-1}=g(z)-e\sum_{n=0}^\infty z^n, $$ and with $g$ being an entire analytic function, as its singularity at $z=1$ is removable. Hence the radius of convergence of the power of $f$ around $z=0$ is the same as the one of $(z-1)^{-1}$, which is equal to $r=1$.
Jonathan Y.
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Yiorgos S. Smyrlis
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