Show series of functions is uniformly convergent on $[a,\infty)$ for $a>0$.

Question

I'd like to show that $$ f(x) = \sum_{n=0}^\infty f_n(x) $$ is uniformly convergent on $[a,\infty)$ for $a>0$ where $$ f_n(x) = \frac{nx}{1+n^4x^2} $$

I have found that $f_n$ has a maximum of $\frac{1}{2n}$ on the interval $[0,\infty)$, specifically at $x=\frac{1}{n^2}$ (except for $f_0(x)=0$ for all $x$). Now this means $$ f\bigg(\frac{1}{n^2}\bigg) = \sum_{n=0}^\infty \frac{1}{2n} \geq \sum_{n=0}^\infty \frac{nx}{1+n^4x^2} \text{ for all }x\in[0,\infty) $$

I am unsure as to where to go from here.

Answer 1

You went in the right direction. Now, as Crostul wrote, you are asked to prove convergence only for $a>0$ so you should consider only the ranks $n\geqslant N$ where $N$ is such that: $a> \frac{1}{N^2}$ so that $f_n(a)$ is the maximum of $f_n$ on $[a,\infty)$.

Then, you are able to prove normal convergence for the series $(\sum f_n)_{n\geqslant N}$ which is just $\sum_{k=0}^{N-1} f_k(x)$ away from $(\sum f_n)_{n\geqslant 0}$ for $x\in [a,\infty)$ since, for $n\geqslant N$:

$$ ||f_n||_\infty = f_n(a) = \frac{na}{1+n^4a^2} = O(\frac{1}{n^3}) $$

as $n\to \infty$.

Thus, $(\sum ||f_n||_\infty)_{n\geqslant N}$ converges since $(\sum \frac{1}{n^3})_{n\geqslant N}$ converges (Riemann series).

Normal convergence is established and it implies uniform convergence.

Answer 2

No need to use Calculus: use the inequality $2nx \leq n^{2}+x^{2}$. $$f_n(x) \leq \frac {n^{2}+x^{2}} {2(1+n^{4}x^{2})} \leq \frac {n^{2}+x^{2}} {2n^{4}x^{2}}=\frac 1 {2n^{2}a^{2}} +\frac 1 {2n^{4}}$$ and the series $\sum \frac 1 {2n^{2}a^{2}}$ and $\frac 1 {2n^{4}}$ both converge.