Finding interval of convergence I need help finishing this problem.
series is $\sum_{n=1}^{\infty}  (1+\frac{1}{n})^{-n^2}e^{-nx}$
I calculated the radius of convergence $\frac{1}{lim_{n\to\infty} ((1+\frac{1}{n})^{-n^2})^{\frac{1}{n}}} = \frac{1}{e}$
I have trouble figuring out the interval of convergence, because there is $e^{-nx}$ instead of (x-x0)^n. I know that $e^{-nx}=(e^{-x})^{n}$
Can someone explain to me how should I proceed?
 A: You have a functions series $\sum_{n}f_{n}(x)$ the power series $\sum_{n}a_{n}x^{n}$ is a particular case of functions series.
We have that
$$|e^{-nx}|\leqslant e^{-n\alpha},\quad \forall x\geqslant \alpha>0$$
Then,
$$\left|\left(1+\frac{1}{n}\right)^{-n^{2}}e^{-nx}\right|\leqslant \left(1+\frac{1}{n}\right)^{-n^{2}}e^{-n\alpha},\quad \forall x\geqslant \alpha$$
Using $n-$th root test, we have
$$\sum_{n=1}^{+\infty}\left(1+\frac{1}{n}\right)^{-n^{2}}e^{-n\alpha}<+\infty.$$
Therefore,
$$\sum_{n=1}^{+\infty}\left(1+\frac{1}{n}\right)^{-n^{2}}e^{-nx}$$ converges uniformly by Weierstrass M-Test for all $x\in [\alpha,+\infty[$.
What can you said about the convergence for the series? Just use $n-$th root test.
A: $a_n=(1+\frac{1}{n})^{-n^2}e^{-nx}$. By root test, if $lim (a_n)^{\frac{1}{n}}$ is smaller than 1 then the series converges. But $(a_n)^{\frac{1}{n}}=\frac{1}{(1+\frac{1}{n})^n(e^x)}\rightarrow \frac{1}{e.e^x}=\frac{1}{e^{x+1}}$. So to have convergence we must have $\frac{1}{e^{x+1}} < 1 \Rightarrow 1 < e^{x+1} \Rightarrow 0<x+1 \Rightarrow -1<x.$
