Determine for each of these series the real numbers $x$ for which it is convergent, and compute the sum for those $x$. $$\sum_{k=0}^\infty (1+x^2)^{-k}\tag a$$ $$\sum_{k=0}^\infty e^{kx}\tag b$$
I compared both of the series with the famous geometric series $$\sum_{n=0}^k a^n = 1+a+a^2+.....+a^k $$ which converges if $|a| < 1 $. So I rewrote (a) to $$ (1/(1+x^2))^k $$ I came to the conclusion that if (a) converges then $$ |a|=\left|1\over1+x^2\right|^k < 1 $$ Am I thinking right? And how do I compute the sum for these $x$?
You're on the right track. Note that the series for $f(x)$ as given by
$$\begin{align} f(x)&=\sum_{k=0}^\infty (1+x^2)^{-k}\\\\ &=\frac{1+x^2}{x^2} \end{align}$$
converges if $\frac{1}{1+x^2}<1$. The $k$ is only a summation "dummy" index and does not appear after carrying out the sum.
Finally, since $\frac{1}{1+x^2}$ is always less than $1$ for $x\ne 0$, the series converges for all $x\ne 0$.
NOTE:
To arrive at the closed-form solution for a geometric series, note that if $S_K(x)=\sum_{k=0}^K x^k$, then $xS_K(x)=\sum_{k=1}^{K+1} x^k$.
Subtracting these expressions, we find $(1-x)S_K(x)=1-x^{K+1}$ whence dividing by $1-x$ (assuming that $x\ne 1$, yields
$$S_K(x)=\frac{1-x^{K+1}}{1-x} \tag 1$$
As $K\to \infty$, the right-hand side of $(1)$ converges if and only if $|x|<1$. Therefore, if $|x|<1$, then
$$\lim_{K\to \infty}\sum_{k=0}^K x^k=\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$