Sum of Geometric Series Formula

Question

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.

Answer 1

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$$S_n=1-\frac1{2^{n+1}}.$$

Answer 2

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 $n = 2$.