Sum of a geometric series of functions: $\sum_{n=0}^\infty\frac{x^2}{(1+x^2)^n}$

Question

$$\text{Let }f_n(x)=\frac{x^2}{(1+x^2)^n} \text{ for } x\in\Bbb{R}$$

$$\text{I found }\sum_{n=0}^\infty f_n(x) = x^2+1 \text{ since it is a geometric series}$$ Now I'm asked to find $a<b$ such that the series converges uniformly on $[a,b]$

I'm inclined to say $a>-\infty$ and $b<\infty$ since for $|x|\ne\infty$, $x^2+1$ is finite. Is this logic sufficient??

Answer 1

Recall: A sequence of functions $g_m\to g$ uniformly on some domain $U$ if and only if $\lim_{m\to\infty}\sup_{x\in U}|g_m(x)-g(x)|=0$.

In this case, let $g_m(x)=\sum_{n=0}^m f_n(x)$ and $g(x)=\sum_{n=0}^\infty f_n(x)=1+x^2$. By straight forward calculation we have $$|g_m(x)-g(x)|=\frac{1}{(1+x^2)^m}$$

Since $\frac{1}{(1+x^2)}$ is continuous on any $[a,b]$, it attains a maximum value $\leq 1$. However, if $[a,b]$ contains $0$, then

$$\sup_{x\in[a,b]}\frac{1}{(1+x^2)}=1$$

and in this case $\lim_{m\to\infty}\sup_{x\in U}|g_m(x)-g(x)|=1$, thus the convergence is not uniform.

If $[a,b]$ does not contain $0$, the convergence is uniform since in this case $$\sup_{x\in[a,b]}\frac{1}{(1+x^2)}<1$$