Is $\sum _{n=1} ^ \infty \frac{1}{(x+\pi)^2 \cdot n^2 }$ uniformly convergent on $(-\pi , \pi)$? 
Is this series uniformly convergent on $(-\pi , \pi)$:
  $$\sum _{n=1} ^ \infty \frac{1}{(x+\pi)^2 \cdot n^2 }\,?$$

My Attempt:
If the series were convergent we would have got a natural number $k$ for a fixed $\epsilon>0$ such that $\sum _{n=k} ^ \infty \frac{1}{(x+\pi)^2 \cdot n^2 } < \epsilon$ for all $x \in (-\pi , \pi)$.  But for this case we will get the term $f_n$ in the summation greater than $1$ for $-\pi + 1/n$.
 So for every $n$ we will get a $x\in (-\pi , \pi) $ such that summation at $x$  is greater than $1$.
That's why the series is not uniformly convergent on $(-\pi , \pi)$.
Can somebody please tell me if I have gone wrong anywhere?
 A: Your answer is correct. The series isn't uniformly  convergent.
A: If a series uniformly converges, then its summands must approach zero uniformly.  
Clearly, the summands do not converge uniformly to $0$.  
To see this, take $\epsilon=1$.  
Then, for all $N$, take any $n>N$ and $x=-\pi +\frac1n$ and find that  
$$\left|\frac{1}{\left(x+\pi\right)^2 n^2}\right|=\left|\frac{1}{\left(-\pi+\frac1n+\pi\right)^2 n^2}\right|=1$$
This negates the uniform convergence of the summand $\frac{1}{(x+\pi)^2n^2}$.  And we are done!

If one does not wish to rely on the aforementioned requirement for a series to converge uniformly, then we can proceed brute force as follows.
With $\epsilon=1$, we have for all $N>1$
$$\begin{align}
\left|\sum_{n=1}^\infty \frac{1}{(x+\pi)^2n^2}-\sum_{n=1}^N \frac{1}{(x+\pi)^2n^2} \right|&=\sum_{n=N+1}^\infty \frac{1}{(x+\pi)^2n^2}\\\\
&\ge \frac1{(x+\pi)^2(N+1)^2}\\\\
&\ge1
\end{align}$$
whenever $x= -\pi+\frac1{ N+1}$.  And we are done!
A: Put $f_n(x)=\displaystyle\sum_{k=1}^n\frac{1}{(x+\pi)^2k^2}$. Then for each $x\in(\pi,\pi)$, $f_n(x)\rightarrow \frac{\pi^2}{6(x+\pi)^2}$. Hence $f_n\xrightarrow{p.w} f$, where $f(x)=\frac{\pi^2}{6(x+\pi)^2}$ for all $x\in(-\pi,\pi)$.
For each $n\in\mathbb{N}$, let 
\begin{eqnarray}
M_n&=&\sup\{|f_n(x)-f(x)|~|~x\in(\pi,\pi)\}\\
 &=& \sup\left\{\left|\displaystyle\sum_{k=n+1}^{\infty}\frac{1}{(x+\pi)^2k^2}\right|~|~~x\in(\pi,\pi)\right\}\\
 &=&\left(\displaystyle\sum_{k=n+1}^{\infty}\frac{1}{k^2}\right)\sup\left\{\left|\displaystyle\frac{1}{(x+\pi)^2}\right|~|~~x\in(\pi,\pi)\right\}\\
&=& \infty
\end{eqnarray}
i.e., $M_n=\infty$ for all $n$ and so $M_n$ does not converges to zero as $n$ tends to $\infty$. Hence $f_n$ does not converges uniformly.
