#for what integer value of x does the series converge?
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|x|<e^2
1\sqrt(R)=lim as n goes to inf (e^(-n)/n)^(1/n)
,w lim as n goes to inf (e^(-n)/n)^(1/n)
There u go homie
best to not just post answers, actually try to help 
From root test, the interval of convergence will be all $x$ such that $\lim_{n\to\infty}\sqrt[n]{\abs{\frac{x^{n/2}e^{-n}}{n}}}<1$. Evaluating the limit, you get $\abs{x}^{1/2}e^{-1}<1$, to which $\abs{x}<e^2$
Omegabet_
@tacit sorrel has given 1 rep to @barren bramble