Find annulus of convergence of Laurent series
$\sum_{-\infty}^{\infty}2^{-n^2}(z-i)^{n^3}$
My answer: $0<|z-i|<\infty$
$\sum_{-\infty}^{\infty}2^{-n^2}(z-i)^{n^2}$
My answer: $|z-i|<\infty$
$\large\sum_{-\infty}^{\infty}\frac{1}{2^{|n|}} z^n$
My answer: $\frac{1}{2}<|z|<2$
$\large\sum_{-\infty}^{\infty}\frac{1}{n^2+1} z^{2n}$
My answer: $0<|z|<\infty$
Can anyone verify those?
Best Answer
Let's deal with the first two series.
In the first series, the general term is $$ \left|2^{-n^2}(z-i)^{n^3}\right|=2^{-n^2+n^3\log_2|z-i|} $$ As $n\to+\infty$ the size of the terms tends to $\infty$ if $|z-i|\gt1$.
As $n\to-\infty$ the size of the terms tends to $\infty$ if $|z-i|\lt1$.
This leaves $|z-i|=1$, for which, the series converges absolutely.
In the second series, the general term is $$ \left|2^{-n^2}(z-i)^{n^2}\right|=2^{-n^2+n^2\log_2|z-i|} $$ As $|n|\to\infty$, the size of the terms tends to $\infty$ if $|z-i|\gt2$.
The size of the terms is $1$ if $|z-i|=2$.
If $|z-i|\lt2$, the series converges absolutely.