I was checking for the uniform convergence of the series $\sum_{n=1}^{\infty} \frac{1}{((x+\pi)n)^2}$ for $x \in (-\pi,\pi)$.
How to think of a bound for $\frac{1}{((x+\pi)n)^2}$ where $x \in (-\pi,\pi)$?, we know $(x+\pi) \in (0,2\pi)$ and for $1\leq(x+\pi)<2\pi $ we have $\frac{1}{((x+\pi)n)^2} < \frac{1}{4\pi^2 n ^2}$ and we are done by the Weierstrass criterion but for $x \in (0,1)$ we have $\frac{1}{((x+\pi)n)^2} > \frac{1}{n^2}$, so its not true for all $x \in (0,2\pi)$ . ?
The true for all $x$ is required as per the definition of Uniform convergence.
Hint. Note that for $x_n=-\pi+1/n$ we have that $$\frac{1}{((x_n+\pi)n)^2}=1.$$ On the other hand, when $x\in [a,\pi)$ for $a\in (-\pi,\pi)$, $$0<\frac{1}{((x+\pi)n)^2}\leq \frac{1/(a+\pi)^2}{n^2}.$$