Series of functions converge uniformly but sequence of functions does not Given $a>1$ and
$$f_{n}(x)=\frac{1}{1+n^{a}x^{4}}$$
I'm asked to show that for any $\delta >0$, the series of functions $\sum f_{n}(x) $ converges uniformly for $\{x \in \mathbb{R} | |x| \geq \delta \}$ but the sequence $\{f_{n}\}$ does not converge uniformly for all real numbers.
A little help...
 A: Hints: 


*

*it converges normally on any $I_\delta \stackrel{\rm def}{=}(-\infty, \delta)\cup(\delta, \infty)$ (for any fixed $\delta > 0$. 
Indeed, for all $n\geq 0$ the function $f_n$ is even, non-negative, and decreasing on $(\delta,\infty)$, so that
$$
\sup_{x\in I_\delta} \lvert f_n(x)\rvert = \sup_{x\in I_\delta} f_n(x) = \frac{1}{1+n^a\delta^4}.
$$
Now, show the series $\sum_n \frac{1}{1+n^a\delta^4}$ is convergent.

*It does not converge uniformly on $\mathbb{R}$, since it does not even converge pointwise. Take $x=0$.

*The sequence $(f_n)_n$ does not converge uniformly (to the indicator function of $0$, $\mathbf{1}_{\{0\}}$, the pointwise limit and therefore the only possible limit) as the uniform limit of a sequence of continuous functions (the $f_n$'s are all continuous) would be continuous. But $\mathbf{1}_{\{0\}}$ is not.
A: For $a>1$ and $x\ge \delta>0$, we have
$$\sum_{n=1}^\infty\frac{1}{1+n^ax^4}\le \sum_{n=1}^\infty\frac{1}{n^a\delta^4}=\frac{1}{\delta^4}\zeta(a)$$
which exists for $a>1$, which one can show using, say, the integral test.
However, we can choose a number $\epsilon=\frac12$, and a number $x=1/n^{a/4}$ such that for any $n$
$$\frac{1}{1+n^ax^4}>\frac12$$
