Does $\sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$ converge uniformly on $(-\infty,\infty)$?

Question

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?

Answer 1

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 $0$.

Answer 2

For (1) you got the idea but the argument is not exactly right.

Indeed the series does converges for all $x$, as you show

$$ \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n} = \begin{cases} 1+ x^2 & \text{if } x\neq 0,\\ 0 & \text{if }x=0.\end{cases}.$$

However it does not converges to a continuous function. Thus the convergence is not uniform (I am using that if $f_n \to f$ uniformly and $f_n$ are continuous for all $n$, then so if $f$.

For (2), it does not converges uniformly to $1+x^2$ on $(-\infty, 0) \cup (0,\infty)$. The set $(-\infty, 0) \cup (0,\infty)$ is not compact.

To check that the convergence is not uniform, write $f_k (x) = \sum_{n=0}^k \frac{x^2}{(1+x^2)^n}$. Then

$$\left| f(x) - f_k(x)\right| = \left| 1+x^2 - x^2 \frac{1-(1+x^2)^{-(k+1)}}{1-(1+x^2)^{-1}}\right| = \frac{1}{(1+x^2)^k}. $$

Thus

$$\sup_{x\neq 0} |f(x) - f_k(x)| = 1$$ for all $k$ and so the convergence is not uniform.