Easy summation question: $S= 1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\cdots$ While during physics I encountered a sum I couldn't evaluate:
$$S= 1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\cdots$$
Is there a particular formula for this sum and does it converges?
 A: Hint: This is a geometric series with ratio $-\dfrac{1}{2}$.
Please see the wikipedia article on geometric series.
In general we have that $1+r+r^2+\ldots=\dfrac{1}{1-r}$ given $|r|<1$.
A: We sum the geometric series:
$$S=\sum_{n=0}^\infty\left(\frac{-1}{2}\right)^n=\frac{1}{1-\left(\frac{-1}{2}\right)}=\frac23$$
A: Another easy way is to note that $2S=2-S$, whence $S=\frac{2}{3}$
A: Using techniques that are frequently used by physicists: \begin{align*}
1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots &= \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{4} - \frac{1}{8} \right) + \cdots \\
&= \frac{1}{2} + \frac{1}{8} + \cdots \\
&= \frac{1}{2} \left( 1 + \frac{1}{4} + \frac{1}{16} + \cdots \right) \\
&= \frac{1}{2} \frac{1}{1-\frac{1}{4}} \\
&= \frac{1}{2-\frac{1}{2}} \\
&= \frac{1}{3/2} \\
&= \frac{2}{3}
\end{align*}
Note:  There is no error in the parenthesization in the first line.  We know the series converges (alternating series test), so every subsequence of the partial sums converges to the same thing, so this particular subsequence of the partial sums converges to the same thing.
A: $$S=1 - \frac {1}{2} + \frac {1}{4} - \frac {1}{8} ...$$ Let this be (1)
Multilply both sides by $\frac {1}{2}$ we get
$$\frac{S}{2}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8} ...$$ and this be (2)
Now Add (1) and (2)
$$\frac{3}{2} \cdot S=1$$
$$ S=\frac{2}{3}$$
