Sum of Geometric Series Formula I just need the formula for the sum of geometric series when each element in the series has the value $1/2^{j+1}$, where $j = 0, 1, 2, \ldots, n$. Please help.
Someone told me it is:
$$S = 2 - \frac{1}{2^n}$$
I am not sure if its right because he has given me no proof and I couldn't prove it when I calculate it manually. Say for example:
$$S = 1/2 + 1/4 + 1/8 = .875$$
But when using the formula given above, with $n=3$ (since there are $3$ elements):
$$S = 2 - 1/8 = 1.875$$
The answers are not the same. Please enlighten me with this issue.
 A: You are computing the sum$$S_n=\frac12+\frac14+\frac18+\cdots+\frac1{2^{n+1}}.$$Well,$$2S_n=1+\frac12+\frac14+\frac18+\cdots+\frac1{2^n}=S_n+1-\frac1{2^{n+1}}$$and therefore$$S_n=1-\frac1{2^{n+1}}.$$
A: Consider the $n$th partial sum
$$S_n = \sum_{j = 0}^{n} r^j = 1 + r + r^2 + \cdots + r^n$$
of the geometric series 
$$\sum_{j = 0}^{\infty} r^j$$
with common ratio $r$.
If we multiply $S_n$ by $1 - r$, we obtain
\begin{align*}
(1 - r)S_n & = (1 - r)(1 + r + r^2 + \cdots + r^n)\\ 
           & = 1 + r + r^2 + \cdots + r^n - (r + r^2 + r^3 + \cdots + r^{n + 1})\\
           & = 1 - r^{n + 1}
\end{align*}
If $r \neq 1$, we may divide by $1 - r$ to obtain
$$S_n = \frac{1 - r^{n + 1}}{1 - r}$$
In particular, if $r = 1/2$, we obtain
\begin{align*}
S_n & = \sum_{r = 0}^{n} \left(\frac{1}{2}\right)^j\\ 
    & = \frac{1 - \left(\frac{1}{2}\right)^{n + 1}}{1 - \frac{1}{2}}\\
    & = \frac{1 - \left(\frac{1}{2}\right)^{n + 1}}{\frac{1}{2}}\\
    & = 2\left[1 - \left(\frac{1}{2}\right)^{n + 1}\right]\\
    & = 2\left(1 - \frac{1}{2^{n + 1}}\right)\\
    & = 2 - \frac{1}{2^n}
\end{align*}
which is the formula you were given.  
However, you want 
\begin{align*}
\sum_{j = 0}^{n + 1} \frac{1}{2^{j + 1}} & = \frac{1}{2} \sum_{j = 0}^{n} \frac{1}{2^j}\\
& = \frac{1}{2} \sum_{j = 0}^{n} \left(\frac{1}{2}\right)^n\\
& = \frac{1}{2}\left[2 - \frac{1}{2^n}\right]\\
& = 1 - \frac{1}{2^{n + 1}}
\end{align*}
As a check, observe that when $n = 2$
$$\sum_{j = 0}^{2} \frac{1}{2^{j + 1}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8} = 0.875$$
and 
$$1 - \frac{1}{2^{2 + 1}} = 1 - \frac{1}{2^3} = 1 - \frac{1}{8} = \frac{7}{8} = 0.875$$
In your calculation, you used $n = 3$ because you did not take into account the fact that if the index starts with $0$, the third term is $n = 2$.
