$\sum_{n=0}^{\infty}\frac{a^2}{(1+a^2)^n}$ converges for all $a\in \mathbb{R}$ $$\sum_{n=0}^{\infty}\frac{a^2}{(1+a^2)^n}$$
Can I just see this series as a geometric series? Since $c = \frac{1}{1+a^2}<1$, we can see this as the geometric series:
$$\sum_{n=0}^{\infty}bc^n = \sum_{n=0}^{\infty}a^2\left(\frac{1}{1+a^2}\right)^n$$
that converges because $c<1$. The sum of this series is:
$$S_n = b(c^0+c^1+c^2+\cdots c^n)\rightarrow cS_n = b(c^1+c^2 + c^3 + \cdots + c^{n}+c^{n+1})\rightarrow $$$$cS_n - S_n = b(c^{n+1}-1)\rightarrow S_n(c-1) = b(c^{n+1}-1)\rightarrow S_n = b\frac{c^{n+1}-1}{c-1}$$
$$S = \lim S_n = b\frac{1}{1-c}$$
So $$\sum_{n=0}^{\infty}\frac{a^2}{(1+a^2)^n} = b\frac{1}{1-c} = a^2\frac{1}{1-\frac{1}{1+a^2}} =$$

Let $a=0$. Then the series obviously converges to $0$. 
Now suppose that $a\ne 0$. Then our series is the geometric series
$$a^2+a^2r+a^2r^2+a^2r^3+\cdots,$$
where $r=\frac{1}{1+a^2}\lt 1$. Since $|r|\lt 1$, the series converges. It is probably by now a familiar fact that when $|r|\lt 1$ the series $1+r+r^2+r^3+\cdots$ converges to $\frac{1}{1-r}$. So our series converges to
$$\frac{a^2}{1-\frac{1}{1+a^2}}.$$
This simplifies to $1+a^2$.
Remark: Your calculation included an almost correct proof that the series that you called $1+c+c^2+c^3+\cdots$ converges to $\frac{1}{1-c}$. There was a little glitch. At one stage you had $\frac{c^{n+1}-1}{c-1}$.  Since $c^{n+1}$ has limit $0$, the sum that you called $S_n$ converges to $\frac{-1}{c-1}$, which I prefer to call $\frac{1}{1-c}$.
I think that part of your calculation is not necessary, since the convergence of $1+c+c^2+\cdots$ to $\frac{1}{1-c}$ if $|c|\lt 1$ can probably by now be viewed as a standard fact.
