For $x \in \mathbb{R}$ consider the series $$ S = \sum_{n=1}^\infty \frac{2x}{x^2+n^2} $$ Then I have to show that $S$ converges uniformly. I have been told before in here that we are not able to use Weiterstrass M-test when the convergent series we find depend on $x$. But how does this change, for example, if we let $x \in [-K,K]$ where $0< K < \infty$? I am studying for my Analysis exam and I have some answers for questions like this where the professor has provided the following answer:
Let $x \in [-K,K]$ where $0 < K < \infty$. Then we have $$ \left| \frac{2x}{x^2+n^2} \right| \leq \frac{2|x|}{n^2} \leq \frac{2K}{n^2} $$ where $\sum_{n=1} \frac{2K}{n^2}$ is well-known convergent. It follows from Weiterstrass M-test that $S$ converges uniformly on $[-K,K]$ where $0 < K < \infty$.
But I simply don't see how this differ from if we just let $x \in \mathbb{R}$ and get that
$$ \left| \frac{2x}{x^2+n^2} \right| \leq \frac{2|x|}{n^2} $$ Why doesn't Weiterstrass M-test work in this instance? It is because $\mathbb{R}$ is not a compact set?
To conclude from M-test that $\sum f_n(x)$ is uniformly convergent you have to produce sequence of positive constants $c_n$ such that $\sum c_n <\infty$ and $|f_n(x)| \leq c_n$ for all $n$ and all $x$. In the present case $\frac {2 |x|} {n^{2}}$ does not have a bound of this type because there is no bound for $|x|$. Hence the bound $\frac {2 |x|} {n^{2}}$ does not yield the conclusion.
EDIT: For uniform convergence on $[-K,K]$ your inequality is good enough.