For which $x \in \mathbb{R}$ is $f(x) = \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}$ continuous/differentiable For which $x \in \mathbb{R}$ the following function is continuous / differentiable:
\begin{align}
f(x) := \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n}
\end{align}
A collection of ideas:
As a hint it is given to use the geometric series. Trying that I found: 


*

*\begin{align}
f(x) := \sum_{n=0}^{\infty} \frac{x^2}{(1+x^2)^n} = x^2 \sum_{n=0}^{\infty} \frac{1}{(1-(-x^2))^n} = x^2+1 
\end{align}
I really don't see how that should help me out! Of course $f(x) = x^2+1$ is continuous and differentiable for all $x \in \mathbb{R}$. 

*The series obviously (follows from 1.) converges for all $x \in \mathbb{R}$ with $x^2 > 0$.

*Based on that I would say the function is continuous and differentiable for all $x \in \mathbb{R}$, but to be honest that appears to be "to simple", so I'm afraid I'm doing something wrong. 
It would be nice if someone could help me to get a clearer picture. 
 A: Note that the geometric series formula $\sum_{n=0}^\infty q^n = \frac1{1-q}$ is applicable only when $|q| < 1$. 
For $x \ne 0$ we have that $\left|\frac1{1 +x^2}\right| < 1$ so using the geometric series gives
$$f(x) = x^2 \sum_{n=0}^\infty \left(\frac{1}{1+x^2}\right)^n = \frac{x^2}{1 - \frac{1}{1+x^2}} = x^2 + 1$$
On the other hand, for $x = 0$ we have $f(0) = 0$.
Therefore
$$f(x) = \begin{cases}
x^2 + 1,  & \text{if $x \ne 0$} \\
0, & \text{if $x = 0$}
\end{cases}$$
so $f$ is continuous and differentiable if and only if $x \ne 0$.
A: It is not a power series, hence for me the notion of radius of convergence has no sense here. However you can use the continuous/derivative theorem for series.
To show it is continuous, you need normal convergence but you need first to find for which $x$ this sum exists. You say $x \in \mathbb{R}$, you have the good feeling but prove it. The series has positive terms and you need to split cases because the behavior of $x^n$ depends on $x>1$, $x<1$, $x=1$


*

*For $|x|>1$
$$
\frac{x^2}{\left(1+x^2\right)^n}\underset{(+\infty)}{\sim}\frac{x^2}{x^{2n}}
$$
Suppose $x \ne 0$, then
$$
\frac{x^2}{\left(1+x^2\right)^n}\underset{(+\infty)}{\sim}\frac{1}{x^{2n-2}}
$$
which converges only for $\left|x\right|>1$.If $x=0$, then it converges.

*If $|x|<1$ 
$$
\frac{x^2}{\left(1+x^2\right)^n}\leq \frac{1}{\left(1+x^2\right)^n}
$$
which converges. Hence no problem.
If $x= \pm 1$ no problem it converges hence it is defined on $\mathbb{R}$. For $x \in \left[a,b\right]$ on $\mathbb{R}^{*+}$ you have
$$
\frac{x^2}{\left(1+x^2\right)^n}\leq \frac{b^2}{1+a^2}
$$ 
Hence, taking the sup, you have normal convergence on every compact set of $\mathbb{R}^{*+}$ and by partity same for $\mathbb{R}^{*-}$ so you have continuity on $\mathbb{R}^{*}$. What about $0$ ?
To show that it is differentiable you need to show that the series of $f'_n$ uniformly converges to its sum.
A: The case $x=0$ is an important subtlety: is a discontinuity since $f(0)$ is the sum of infinitely many zeroes.
