Finding the sum to infinity Question:
Find the sum to infinity for the following series
$$1, -\frac{1}{2}, \frac{1}{2^2}, -\frac{1}{2^3},\cdots$$
What would be the technique used to find such a sum?
 A: Let
$\text {eq.}{(1)}$ be,$$S=1 -\frac{1}{2}+ \frac{1}{2^2}-\frac{1}{2^3}+\cdots\infty$$
Then,
$\text {eq.}{(2)}$ be
$$\frac{-S}2=-\frac{1}{2}+ \frac{1}{2^2}-\frac{1}{2^3}+\cdots\infty$$
Subtract 2 from 1
$$\frac{3S}2=1$$
Or, $$S=\frac{2}3$$
A: Perhaps a bit simpler:
The $n$'th term of your series (assuming we start at $n=0$) can be written as
$$\frac{(-1)^n}{2^n}$$
which is just $(-1/2)^n$. Since $|-1/2| < 1$, it's a convergent geometric series of the form
$$\sum_{n=0}^{\infty} x^n$$
where $x = -1/2$. The series converges to
$$\frac{1}{1-x} = \frac{1}{1 + \frac{1}{2}} = \frac{2}{3}$$
A: Write the sum of the $n$ and $(n+1)^{th}$ terms as 
$$
  \frac{1}{2^n} - \frac{1}{2^{n+1}}
= \frac{1}{2^{n+1}}
$$
then sum them up to get
$$
  \sum_{n \in 2 \mathbb{N} }^\infty \frac{1}{2^{n+1}}
= \sum_{k=0}^\infty \frac{1}{2^{2k+1}}
= \frac{1}{2} \sum_{k=0}^\infty \frac{1}{4^k}
= 2/3.
$$
A: $(1+\frac{1}{2^2}+\frac{1}{2^4}+...)-(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+...)$
Both the brackets above contain GP of $n$ terms with $r=\frac{1}{2^2}=\frac{1}{4}$
$\Large(\frac{(\frac{1}{4})^n-1}{-3/4})-(\frac{1}{2}\frac{(\frac{1}{4})^n-1}{-3/4})$
$\Large\frac{\frac{1}{2}(\frac{1}{4})^n-\frac{1}{2}}{\frac{-3}{4}}=\frac{(\frac{1}{4})^n-1}{\frac{-3}{2}}=\frac{0-1}{\frac{-3}{2}}=\frac{2}{3}$
A: Here is another way, given that the series converges:
Let $S$ be the sum, then note that $S = 1 - {1 \over 2} S$, solving gives the desired result.
Note:
To see where the above comes from, note that $S=1-{1 \over 2} + {1 \over 2^2}\cdots$, and so
$-{1 \over 2} S = -{1 \over 2} + {1 \over 2^2}\cdots$. Hence we have the equation $S = 1 - {1 \over 2} S$.
(This is contingent on the series being convergent, otherwise one can end up with nonsense.)
A: consider the sum as x. Then 4x = (4 - 2) + 1 - 1/2 ...
So 4x = (2) + x
Then x = 2/3
