Consider $$\sum_{n=1}^\infty \frac{1}{n^2(1 + x/n)}$$
where $x \in (-1,\infty)$.
At $n = 1$, we have $\frac{1}{1 + x}$, and surely this can be made as close to infinity as we want. Is it then not possible to use Weierstrass' M-test to show uniform convergence?
I'm still trying to get my head around this topic, so I might be missing something obvious.
$$\sum_{n=1}^{+\infty}\frac{1}{n^2(1+x/n)}=\sum_{n\geq 1}\frac{1}{n(x+n)}=\frac{\gamma+\psi(x+1)}{x}\tag{1}$$ where $\psi(x)=\frac{d}{dx}\log\Gamma(x)$ is the digamma function and $\gamma$ is the Euler-Mascheroni constant.
The regularity of the RHS of $(1)$ over $x>-1$ can be deduced from the regularity of $\Gamma(s)$ over $\text{Re}(s)>0$ or from the Cauchy-Schwarz inequality, from which:
$$\left|\sum_{n\geq 1}\frac{1}{n(x+n)}\right|\leq \sqrt{\zeta(2)}\sqrt{\sum_{n\geq 1}\frac{1}{(x+n)^2}}.\tag{2}$$