Which of the following series is uniformly convergent?

Question

For $x\in (-\pi, \pi) $, I want to know whether the following series following series are uniformly convergent on $(-\pi,\pi) $:
1. $\sum _{n=1}^{\infty}{(\frac {x}{n})^n} $
2. $\sum_{n=1}^{\infty}{\frac {1}{((x+\pi)n)^2}}$.
I think series in 2 has to be uniformly convergent for,
$|\frac{1}{((x+\pi)n)^2}|$ $\leq $ $\frac {1}{n^2} $ and as $\sum {\frac {1}{n^2}} $ is convergent, by Wierstrass M-test the series in (2) is uniformly convergent.
But I don't how to proceed for the series in (1). Any help please! Also tell if my answer to the (2)nd is correct.

Answer 1

For part (1) you have $$ \left|\frac{x}{n}\right|^n< \left|\frac{\pi}{n}\right|^n$$ and $$\sum_{n=1}^\infty \left|\frac{\pi}{n}\right|^n$$ converges so the M-test applies here.

Part (2) is funny, cause it seems pretty benign: $$ \sum_{n=1}^\infty \frac{1}{((x+\pi)n)^2} = \frac{1}{(x+\pi)^2}\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6(x+\pi)}$$ and yet it doesn't uniformly converge since whatever small difference remains from the nicely converging $\sum \frac{1}{n^2}$ is blown up arbitrarily large by the divergence as $x\to -\pi$ (as user284331 shows in detail in their answer).

Answer 2

Number two is not correct: Let $S_{n}(x)=\displaystyle\sum_{k=1}^{n}\dfrac{1}{(x+\pi)^{2}}\dfrac{1}{k^{2}}=\dfrac{1}{(x+\pi)^{2}}\displaystyle \sum_{k=1}^{n}\dfrac{1}{k^{2}}$, $S(x)=\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{(x+\pi)^{2}}\dfrac{1}{k^{2}}=\dfrac{1}{(x+\pi)^{2}}\displaystyle \sum_{k=1}^{\infty}\dfrac{1}{k^{2}}$, then $|S_{n}(x)-S(x)|=\dfrac{1}{(x+\pi)^{2}}\displaystyle \sum_{k\geq n+1}\dfrac{1}{k^{2}}$, if it were uniformly convergent, then for some $N$, we have $|S_{n}(x)-S(x)|<1$ for all $n\geq N$ and $x\in(-\pi,\pi)$, then fix the $n$, take $x\rightarrow-\pi^{+}$ and the expression blows up.

Compliment to @spaceisdarkgreen answer: $\left|\dfrac{\pi}{n}\right|^{n}\leq\left(\dfrac{n^{1/2}}{n}\right)^{n}=\dfrac{1}{n^{n/2}}\leq\dfrac{1}{n^{2}}$ for large $n$, so the series converges.