Q:

How to find region of convergence of the following series? 

Find the region of convergence of the following series$$\sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)^{-\frac{n^2}{4}}\cdot(x-1)^{\frac{n}{3}}$$

I have done: $$\left(1+\frac{1}{n}\right)^{n\left(-\frac{n}{4}\right)}\sim{e^{-\frac{n}{4}}}$$But what is the next step to find region of convergence? 

A:

Hint.  Consider the limit of
$$L:=\lim_{n\to \infty}|a_n|^{1/n}=\lim_{n\to \infty}(1+\frac{1}{n})^{-\frac{n}{4}}|x-1|^{\frac{1}{3}}=\frac{|x-1|^{\frac{1}{3}}}{e^{1/4}}.$$
What happens when $L<1$ or $L>1$?  What about the case when $L=1$?  
P.S. Remember that when $L=1$ a more detailed analysis is needed since the root-test is inconclusive.  We have that $|x-1|^{\frac{1}{3}}=e^{1/4}$ and 
$$|a_n|=e^{n/4}(1+\frac{1}{n})^{-\frac{n^2}{4}}=\left(\frac{e^n}{(1+\frac{1}{n})^{n^2}}\right)^{1/4}\to e^{1/8}\not=0$$
where in the last step a Taylor expansion (or Hopital) is needed.  Hence the series is not convergent when $L=1$.

