# How do you find the explicit formula for the following sequence 1, 1/2, 1/4, 1/8,...?

Mar 9, 2016

${n}^{t h}$ term $\frac{1}{2} ^ \left(n - 1\right)$ and sum up to $n$ terms is $2 \left(1 - \frac{1}{2} ^ n\right)$

#### Explanation:

The sequence $\left\{1 , \frac{1}{2} , \frac{1}{4} , \frac{1}{8} , . .\right\}$ is a geometric series of the type $\left\{a , a , a {r}^{2} , a {r}^{3} , \ldots .\right\}$, in which $a$ - the first term is $1$ and ratio $r$ between a term and its preceding term is $\frac{1}{2}$.

As the ${n}^{t h}$ term and sum up to $n$ terms of the series $\left\{a , a , a {r}^{2} , a {r}^{3} , \ldots .\right\}$ is $a {r}^{n - 1}$ and $\frac{a \left(1 - {r}^{n}\right)}{1 - r}$ (as $r < 1$ - in case $r > 1$ one can write it as $\frac{a \left({r}^{n} - 1\right)}{r - 1}$.

As such ${n}^{t h}$ term of the given series $\left\{1 , \frac{1}{2} , \frac{1}{4} , \frac{1}{8} , . .\right\}$ is $1 \times {\left(\frac{1}{2}\right)}^{n - 1}$ or $\frac{1}{2} ^ \left(n - 1\right)$

and sum up to $n$ terms of the series is $\frac{1 \times \left(1 - {\left(\frac{1}{2}\right)}^{n}\right)}{1 - \frac{1}{2}}$ or

$\frac{\left(1 - \frac{1}{2} ^ n\right)}{\frac{1}{2}}$ or $2 \left(1 - \frac{1}{2} ^ n\right)$.

Note that when $n \to \infty$, $\frac{1}{2} ^ n \to 0$ and hence sum of the series tends to $2$.