[Math] Partial sum of a geometric series

Question

Consider the geometric series $\sum_{n=0}^\infty ar^n$ where $a=1$
and $r=-\frac{1}{2}$. Since $|r| < 1$, the series converges to
$S = \sum_{n=0}^\infty ar^n = \frac{1}{1+\frac{1}{2}} = \frac{2}{3}$.

I would like to arrive at the same sum by computing
$\lim_{N \to \infty} S_N$ where $S_N$ is the partial sum of $N$ terms of
the goemetric series.

First few terms of the geometric series are: $1, -\frac{1}{2}, \frac{1}{4},
-\frac{1}{8}, \frac{1}{16}, -\frac{1}{32}, \cdots$.

Here are my attempts to find $S_N$:
\begin{align*}
S_1 &= 1 = \frac{2^{(1-1)}}{2^{(1-1)}} \\
S_2 &= 1 – \frac{1}{2} = \frac{1}{2} = \frac{2^{(2-1)} – 1}{2^{(2-1)}} \\
S_3 &= \frac{1}{2} + \frac{1}{4} = \frac{3}{4} =
\frac{2^{(3-1)} – 1}{2^{(3-1)}} \\
S_4 &= \frac{3}{4} – \frac{1}{8} = \frac{5}{8} =
\frac{2^{(4-1)} – 3}{2^{(4-1)}} \\
\cdots \\
S_N &= ??
\end{align*}
\begin{align*}
S_1 &= 1 – \frac{0}{1} \\
S_2 &= 1 – \frac{1}{2} \\
S_3 &= 1 – \frac{1}{2} + \frac{1}{4} = 1 – \frac{1}{4} \\
S_4 &= 1 – \frac{1}{4} – \frac{1}{8} = 1 – \frac{3}{8} \\
\cdots \\
S_N &= ??
\end{align*}

I cannot find a pattern that will help me find $S_N$.

Answer 1

Hint:

$$a + ar + ar^2 + \ldots + ar^{n-1} = \frac{ a( 1- r^n) } { (1-r) }. $$