206.10.3.17 Evaluate the following geometric sum

[karush]

$\tiny{206.10.3.17}$
$\textsf{Evaluate the following geometric sum.}$
$$\displaystyle
S_n=\frac{1}{2}+ \frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\cdots + \frac{1}{8192}$$
$\textsf{This becomes}$
$$\displaystyle
S_n=\sum_{n=1}^{\infty}\frac{1}{2^{2n-1}}=\frac{2}{3}$$
$\textsf{How is this morphed into the geometric formula?}$

 

[Prove It]

$\tiny{206.10.3.17}$
$\textsf{Evaluate the following geometric sum.}$
$$\displaystyle
S_n=\frac{1}{2}+ \frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\cdots + \frac{1}{8192}$$
$\textsf{This becomes}$
$$\displaystyle
S_n=\sum_{n=1}^{\infty}\frac{1}{2^{2n-1}}=\frac{2}{3}$$
$\textsf{How is this morphed into the geometric formula?}$

Notice that it can be written as

$\displaystyle \begin{align*} \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \cdots &= \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} + \frac{1}{128} + \frac{1}{256} + \dots \right) - \left( \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \dots \right) \\ &= \sum_{n = 1}^{\infty}{ \left( \frac{1}{2} \right) ^n } - \sum_{n = 1}^{\infty}{ \left( \frac{1}{4} \right) ^n } \\ &= \frac{\frac{1}{2}}{1 - \frac{1}{2}} - \frac{\frac{1}{4}}{1 - \frac{1}{4}} \\ &= \frac{\frac{1}{2}}{\frac{1}{2}} - \frac{\frac{1}{4}}{\frac{3}{4}} \\ &= 1 - \frac{1}{3} \\ &= \frac{2}{3} \end{align*}$

 

[karush]

$\text{how is $\displaystyle a=\frac{1}{2}$ and $\displaystyle r=\frac{1}{2}$ derived from that?}$

 

[MarkFL]

\(\displaystyle S=\sum_{k=1}^{\infty}\left(\frac{1}{2^{2n-1}}\right)=\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{1}{2^{2n-2}}\right)=\frac{1}{2}\sum_{k=1}^{\infty}\left(\left(\frac{1}{4}\right)^{n-1}\right)=\frac{1}{2}\cdot\frac{1}{1-\dfrac{1}{4}}=\frac{1}{2}\cdot\frac{4}{3}=\frac{2}{3}\)

 

[MarkFL]

Another way to evaluate this series is to observe that, in binary, it is the repeating decimal:

\(\displaystyle S=0.\overline{10}=\frac{10}{11}\)

Converting to decimal, this becomes:

\(\displaystyle S=\frac{2}{3}\)

 

[alyafey22]

$$S_n=\frac{1}{2}+ \frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\cdots + \frac{1}{8192}$$

The right hand side should be as a function of $n$.

 

[Greg]

$\tiny{206.10.3.17}$
$\textsf{Evaluate the following geometric sum.}$
$$\displaystyle
S_n=\frac{1}{2}+ \frac{1}{8}+\frac{1}{32}+\frac{1}{128}+\cdots + \frac{1}{8192}$$
$\textsf{This becomes}$
$$\displaystyle
S_n=\sum_{n=1}^{\infty}\frac{1}{2^{2n-1}}=\frac{2}{3}$$
$\textsf{How is this morphed into the geometric formula?}$

$$\sum_{n=1}^{\infty}\frac{1}{2^{2n-1}}=2\sum_{n=1}^{\infty}\frac{1}{2^{2n}}=2\sum_{n=1}^{\infty}\frac{1}{4^n}=2\left(\frac{\frac14}{1-\frac14}\right)=\frac23$$