Is $f(x)=x^2+\frac{x^2}{1+x^2}+\frac{x^2}{(1+x^2)^2}+\dots,x \in \Bbb{R}$ continuos? 
Problem Is $f$ continuous? Where,  $f(x)=x^2+\frac{x^2}{1+x^2}+\frac{x^2}{(1+x^2)^2}+\dots,x \in \Bbb{R}$.

My Solution. Let $f_n(x)=\frac{x^2}{(1+x^2)^n},x \in \Bbb{R}$. I've checked the convergency of the series $\sum f_n$:
Now $S_n(x)=\sum_{k=0}^{n}f_n(x)=x^2[1+\frac{1}{1+x^2}+\dots+\frac{1}{(1+x^2)^{n}}]=(1+x^2)[1-\frac{1}{(1+x^2)^{n+1}}],x \in \Bbb{R}$
Now for any $x \in \Bbb{R}, \lim_{n\to \infty} S_n(x)=1+x^2(=g(x),say)$
Therefore $S_n \to g$ pointwise on $\Bbb{R}$. Hence $f(x)=g(x)$ for all $x\in \Bbb{R}$. 
Can I say now that $f$ is a continuous function on $\Bbb{R}$ as $f(x)=g(x)$ for all $x$?
But as we can see $S_n$ is not uniformly convergent to $g$ .....Since $\|S_n-g\|_{\infty}\ge 1$...!!!
I am getting confused here....$f$ is pointwise equal to a continuous function ON THE OTHER HAND the series $\sum f_n$ is not uniformly convergent on $\Bbb{R}$...!!!
How to decide whether $f$ is continuous or not....?? Please tell me where I made a mistake..Thank you.

As I commented, $f(0) \neq g( 0)$, since the formula of geometric series is not applicable when $x = 0$. Thus 
$$
f(x) = 
\begin{cases}
0, & x=0\\
g(x), & \text {else}
\end{cases}. 
$$
