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What kind of singularity does this function have:

$$\sum_{n=0}^\infty \frac{1}{z^{n}n!}.$$

It can have pole but its answer is still zero after multiplication by $z^n$ at $n=0$. Therefore the second choice is that it has an essential singularity. Is that correct?

GaussTheBauss
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    The way I like to think about it is that if a Laurent expansion has infinitely many negative terms, then it has an essential singularity. – Cameron Williams Dec 26 '15 at 04:20

1 Answers1

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Yes, this has an essential singularity at $0$: For each $k$, the function $z^k \sum_{n = 0}^{\infty} \frac{1}{z^n n!}$ is unbounded at zero, so the function has a singularity that is not a pole of order $k$. Alternatively, you've already got the full Laurent series: The series coincides with the essential part and is non-terminating.

Note that we can explicitly compute the sum:

$$\sum_{n = 0}^{\infty} \frac{1}{z^n n!} = \sum_{n = 0}^{\infty} \frac{(1/z)^n}{n!} = e^{1/z}$$

Now knowing that $e^z$ has an essential singularity at infinity translates into an essential singularity at $0$ for this function.

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