Integrate $\int_{|z|=1}z^{3}e^{1/z}dz$ - verification

Question

I integrate over a circular path centered at 0 with radius 1

$\int_{|z|=1}z^{3}e^{1/z}dz=\int_{|z|=1}z^{3}\sum\limits_{n=0}^{\infty}\frac{1}{n!z^{n}}dz=\int_{|z|=1}\sum\limits_{n=0}^{\infty}\frac{1}{n!z^{n-3}}dz$

Then we can switch integration and summation.

$\int_{|z|=1}\sum\limits_{n=0}^{\infty}\frac{1}{n!z^{n-3}}dz=\sum\limits_{n=0}^{\infty}\frac{1}{n!}\int_{|z|=1}\frac{1}{z^{n-3}}dz=\frac{1}{2!}\int_{|z|=1}\frac{1}{z}dz=\pi i$

Since for a circular path around $0$ integrals of form

$\int_{|z|=1}\frac{1}{z^n}dz$

are $0$ for all $n\neq1$ and $2\pi i$ for $n=1$

Answer 1

It's almost correct. In your summation the $z^{-1}=1/z$ term comes at $n=3+1=4$, not at $n=3-1=2$. You should have gotten $i\pi/12$.

Watch your signs!

Answer 2

It is correct. In my opinion, you should add that your series conveges uniformly in $S^1$. Otherwise, you cannot know whether you can exchange the series and the integral.