Proof verification: integral of $\sum\limits_{n=1}^{\infty} \frac{1}{n! z^n}$ over the unit circle

Question

Given the series $\displaystyle{\sum_{n=1}^{\infty}} \frac{1}{n! z^n}$ I have to show that it defines a holomorphic function in $\mathbb{C}-\{ 0 \}$ and then I have to calculate the integral of this function over the unit circle.

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For the first point, it is sufficient to note that this is the Laurent series of a function which is holomorphic in the annulus of convergence. A rapid calculation of the $\text{lim} \frac{1}{n!}^{\frac{1}{n}}=0$ shows that the series is convergent in all of $\mathbb{C}$ without $0$ obviously.

Now, the function defined by this series is a function which is holomorphic in the domain of existence, but has a singularity in $0$, the nature of the singularity is essential, because the Laurent series of the function is the series given above which has a principal part which is not finite in number of coefficients. So I cannot utilise the residue formula for the integral.

But i noticed that the series is uniformly convergent in all the circle, so i can integrate term by term obtaining (for the generic $n$th-term) the following integral:

$$-\frac{1}{n!(n-1)} \int_{0}^{2\pi} \frac{\text{d}}{\text{d}\theta} e^{-i\theta(n-1)} \text{d}\theta$$

which is $0$ for every $n \in \mathbb{N}$. So that I can conclude that the integral is null.

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Is my proof correct?

Answer 1

You wrote that what you have is a Laurent series, and you are right. So, you know that the residue of your function at $0$ is the coefficient of $\frac1z$, which is $1$. So, by the residue theorem, your integral is equal to $2\pi i$.

Your computation is obviously wrong, because you divide by $n-1$. What does that mean if $n=1$? If your series was $\displaystyle\sum_{n=2}^\infty\frac1{n!z^n}$, then you would be correct.

Answer 2

I understood my error.

First: I can actually use the residue formula, but I knew that I can utilise it only for poles, so to follow my proof and find the error, for now I will ignore the residue formula.

The flaw in my proof is that I didn't isolate the $n=1$ case (as @José pointed out).

Indeed if one consider the original term by term integral, the general $n$th term is:

$$\frac{i}{n!} \int_{0}^{2\pi} e^{-i\theta (n-1)}\text{d}\theta $$

For $n>1$, because of my reasoning above, follows that every integral is $0$ (this is like the proof of the residue formula actually).

While, for $n=1$ we have that the integrand is $1$ and so follows that the integral is $2\pi i $ and the result follows.