easy Question about uniform convergence I have a fairly easy question of uniform convergence, I believe.
I have function: $$\sum\frac{1}{n}-\frac{1}{x^2+n}=\sum \frac{x^2}{nx^2+n^2}\le1$$
Why is this only uniformly convergent on finite intervals rather than all reals?
 A: *

*In case of finite interval: 
$\exists~k>0$ such that $|x|<k~\forall~x$ in the domain. Consequently , 
$$\left|\dfrac{x^2}{nx^2+n^2}\right|\leq\left|\dfrac{x^2}{n^2}\right|<\dfrac{k^2}{n^2}.$$ Since $\sum \dfrac{k^2}{n^2}$ is convergent, by Weierstrass M-test $\sum \dfrac{x^2}{nx^2+n^2}$ is uniformly convergent in the domain.

*In case of $\mathbb R:$ 
$\forall~n\in\mathbb N,~f_n(x)=\dfrac{x^2}{nx^2+n^2}\to 1$ as $x\to\infty.$ So for $\epsilon=\dfrac{1}{2},$ there's no $k\in\mathbb N$ such that $$|f_{k+1}(x)+f_{k+2}(x)+...+f_{k+p}(x)|<\dfrac{1}{2}~\forall~p\in\mathbb N ~\text{and} ~\forall~x\in\mathbb R.$$
In fact it fails to hold for $p=1$ whence by Cauchy's principle of convergence $\sum \dfrac{x^2}{nx^2+n^2}$ is not uniformly convergent in $\mathbb R$.
A: It's because $\displaystyle\lim_{x \to \infty} \frac{x^2}{nx^2 + n^2} = 1$ for any $n$. So while there may be some $N$ guaranteeing $\epsilon$-uniform convergence for some bounded interval, we can choose $x$ large enough so that $\displaystyle \frac{x^2}{Nx^2 + N^2} \approx 1$ if we don't bound $x$.
