Convergence and Unifrom convergence of $\sum_{n=1}^{\infty} (\frac{x}{n})^n$

Question

Is $\sum_{n=1}^{\infty}{(\frac{1}{n})^n}$

1)convergent aand 2) uniformly convergent?

where $x \in (-\pi,\pi)$

I was thinking to use Weierstrass criterion to show uniform convergence by getting an upper bound which converges,

$\frac{x}{n} < \frac{\pi}{n}$

$\frac{1}{n}^n < \frac{1}{n}$.

$(\frac{x}{n})^n<(\frac{\pi}{n})^n < \pi ^n . \frac{1}{n}$

But i could not find the bound?.Any other way I can proceed?

Answer 1

$|\frac x n|^{n} \leq |\frac {\pi} n|^{n} \leq (\frac 1 2 )^{n}$ for $n$ sufficiently large. Compare with $\sum (\frac 1 2 )^{n}$

Answer 2

It is a power series where $a_n=1/n^n$ and $\operatorname{limsup}_{n\to \infty}|a_n|^{1/n}=\operatorname{limsup}(1/n)=0$,so the radius of convergence is $\infty$ i.e. it converges on entire $\mathbb R$.Now it will converge on any compact interval in $\mathbb R$.But I think not on entire real line.