For $x \in (-\infty; -2]\cup [0;+\infty): \frac{1}{n^2 (x+1)}\le \frac{1}{n^2} $, so it conv. uniformly on $(-\infty; -2]\cup [0;+\infty)$ by the Weierstrass M-Test.
The problem is what happens when $x\in (-2;0)$? I tried using Cauchy criterion and choosing $x = -1 +\frac1{n^2} $, so that $|\sum_{k = m+1}^{n} \frac{1}{n^2 (x+1)}|\ge \epsilon$ for some given $\epsilon > 0$. But this can only show that the series is not uniformly convergent on the entire $(-2;0)$. What about any subinterval of it? Does it not converge on any subinterval? And how to prove it?
Where exactly does the series of functions $\sum_{n = 1}^{\infty} \frac{1}{n^2 (x+1)}$ converge uniformly
real-analysissequences-and-seriesuniform-convergence
Best Answer
$\frac 1 {n^{2}(x+1)}$ does not tend to $0$ uniformly in any open interval containing or touching $-1$. (To see this take $x=-1\pm \frac 1 {n^{3}}$). So the series does not converge uniformly in any such interval, but it converges uniformly in any interval that stays away from $-1$ (by M-test)