<s>[INST] Proving uniform convergence 
Prove uniform convergence of function series: $$ \sum_{n=0}^\infty \frac{1}{n^2  + x} \sin \frac{1}{n^2  + x}$$ on $ \Bbb R $

I'm stuck with a problem, because I've proven that is uniformly convergent(proof below), while a friend of mine, gave a counterexample to this problem, $$ x = \frac{1 - \frac{\pi n^2}{2}}{\pi/2} $$ and then it is only $$\sum_{n=0}^\infty \frac{\pi}{2}\sin \frac{\pi}{2} =  \sum_{n=0}^\infty \frac{\pi}{2} $$ , so it not uniformly convergent.
Here is my proof of uniform convergence:
$$ \frac{1}{n^2  + x} \sin \frac{1}{n^2  + x} <  \frac{1}{n^2  + x}    $$ , since $$ \sin {x} < 1 $$ 
and 
$$ \frac{1}{n^2  + x}  < \frac{1}{n^2} $$ , which is convergent.
Hence, I claim that by Weierstrass M-Test : $$ \sum_{n=0}^\infty \frac{1}{n^2  + x} \sin \frac{1}{n^2  + x} $$ is uniformly convergent on $$ \Bbb R $$ . 
So, where is the mistake? I'd appreciate any help.
P.S Sorry, for any mistakes in latex.
EDIT: I'm awfully sorry it should be : $$ \sum_{n=0}^\infty \frac{1}{n^2  + x} \sin \frac{1}{n^2  + x}$$
 [/INST] As $x$ approaches $-n^2$, $\frac1{n^2+x}\sin\left(\frac1{n^2+x}\right)$ oscillates wildly between very large numbers. Since uniform convergence only worries about the tail, we can fix this by assuming that $x$ is not the negative of a perfect square and that $x\gt-m^2$.
Let $M,N\gt2m$, then
$$
\begin{align}
\sum_{n=M}^N\frac1{n^2+x}
&\le\sum_{n=M}^N\frac1{n^2-m^2}\\
&\le\sum_{n=M}^N\frac1{n^2-(n/2)^2}\\
&\le\frac43\sum_{n=M}^N\frac1{n^2}\\
&\le\frac43\sum_{n=M}^N\frac1{n(n-1)}\\
&=\frac43\left(\frac1{M-1}-\frac1N\right)
\end{align}
$$
This is independent of $x$ as long as $x\gt-m^2$.
</s>