Does $\displaystyle \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$
converge uniformly on $(-\infty,\infty)$?
My attempt:
No. Consider the case where $x=0$, then
$\displaystyle \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n} = 0$.
For $x \neq 0$, observe $\displaystyle 0 \lt \frac{1}{(1+x^2)^n} \lt 1$, so by geometric series formula
$\displaystyle \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$
$\displaystyle = \frac{x^2}{1 – \frac{1}{1+x^2}} = 1+x^2$
(1) So clearly the series doesn't even converge for all $x$, let alone converge uniformly.
Now, my question is about the case where $x \neq 0$. Does it converge uniformly to $1 + x^2$?
(2) I think, "yes". By Dini's theorem for series the convergence of the series to $1 + x^2$ must be uniform since $1+x^2$ is continuous and $(-\infty,0) \cup (0,\infty)$ is compact.
Is my reasoning for (1) and (2) correct?
Best Answer
What you write under (1) is incorrect. You proved that the series converges for $x=0$, and for any fixed $x\neq 0$. So the series converges pointwise.
The argument against uniform convergence on $(-\infty,\infty)$ then is immediate: letting $f$ be the (pointwise) limit of the series, you have $f(0)= 0$ but $f(x) = 1+x^2$ if $x\neq 0$. Well, if the series was uniformly convergent, since it's the sum of continuous functions, the limit $f$ would be continuous... it clearly isn't at $0$.