convergence of $\sum_{n=1}^{\infty} \frac{e^{-xn\ln x}}{x^2 + n}$ if $x>0$

By substitution $a=x^{-x}$ for $x>1$ you obtain a series $\sum_{n=1}^{\infty}\frac{a^n}{b+n},~0<a<1,~b>0$. Using the ratio test you immediately obtain convergence: $\left|\frac{\frac{a^{n+1}}{b+n+1}}{\frac{a^n}{b+1}}\right|=a\left|\frac{b+n}{b+n+1}\right|<1$. For $x=1$ you obtain a harmonic series, which is known to diverge. And finally for $0<x<1$, for which $a=x^{-x}>1$, by using the ratio test again you can find sufficiently high $n$ that the ratio is greater than $1$, ergo the given series diverges as well. So the series converges for $x>1$, otherwise diverges.

I appear to get a different answer. $$ a_n = \frac{x^{-nx}}{x^2+n} $$ $$\implies \lim_\limits{n \to \infty} (a_n)^{\frac{1}{n}} = \lim_\limits{n \to \infty} \frac{x^{-x}}{(x^2+n)^\frac{1}{n}} $$ $$ = \lim_\limits{n \to \infty} \frac{x^{-x}}{x^{\frac{2}{n}}(1+\frac{n}{x^2})^\frac{1}{n}} $$ Now $x^{\frac{2}{n}} \to 1$ and $ \Big(1+\frac{n}{x^2}\Big)^\frac{1}{n} \to e^{\frac{n}{nx^2}} = e^{\frac{1}{x^2}} $

$$ \implies \lim_\limits{n \to \infty} (a_n)^{\frac{1}{n}} = x^{-x} e^{\frac{1}{x^2}} $$

Plugging this in a graphing calculator (or equivalently analyzing using derivatives) shows that its always lesser than 1. So, shouldn't the series converge $\forall x \in \mathbb{R}^+$