Geometric Series

Idea. A geometric series is the simplest series: we know exactly when it converges and what it sums to. The only work is to identify the ratio and check the condition $|r|<1$.

First series. Here $r=-\frac12$, and $|r|=\frac12<1$, so the series converges with sum
$$\frac{1}{1-(-1/2)}=\frac{1}{3/2}=\frac23.$$

Second series. Here $r=\frac{a}{1+a}$. The convergence condition $|r|<1$ means
$$\left|\frac{a}{1+a}\right|<1.$$
Since $1+a$ must be nonzero (otherwise $r$ is undefined), we solve: $|a|<|1+a|$. Squaring both sides, $a^2<(1+a)^2=1+2a+a^2$, so $0<1+2a$, i.e. $a>-\frac12$. Under this condition the sum is
$$\frac{1}{1-\frac{a}{1+a}}=1+a.$$

Third series. Here $r=1-x^2$. Convergence requires $|1-x^2|<1$, i.e. $-1<1-x^2<1$. The left inequality gives $x^2<2$, the right gives $x^2>0$. So the series converges exactly for $x\in\,]0,\sqrt2[\,\cup\,]-\sqrt2,0[$. Its sum is
$$\frac{1}{1-(1-x^2)}=\frac{1}{x^2}.$$

Remark. In all three cases the answer is a *function* of the parameter: the sum of a geometric series depends continuously on $r$ on the interval $|r|<1$, and blows up as $r\to1^-$.