# How do you find the sum of 1+1/5+1/7^2+1/7^3+1/7^9+...+1/7^n+...?

Feb 2, 2017

$1 + \frac{1}{5} + \frac{1}{7} ^ 2 + \frac{1}{7} ^ 3 + \frac{1}{7} ^ 9 + \ldots + \frac{1}{7} ^ n + \ldots = 1 + \frac{1}{5} + \frac{1}{49} + \frac{1}{343} + \frac{7}{6} - \frac{282475242}{242121642} \cong 1.223323644$

#### Explanation:

The question is not very clear...in answering I assume that for $n \ge 9$ the sum is on all values of $n$

We can write the sum as:

$1 + \frac{1}{5} + \frac{1}{49} + \frac{1}{343} + {\sum}_{n = 9}^{\infty} \frac{1}{7} ^ n$

Focus on the series: clearly we have:

${\sum}_{n = 9}^{\infty} \frac{1}{7} ^ n = {\sum}_{n = 0}^{\infty} \frac{1}{7} ^ n - {\sum}_{n = 0}^{8} \frac{1}{7} ^ n$

Thus at the second member we have the sum of a geometric series and the partial sum of a geometric series of ratio $\frac{1}{7}$:

${\sum}_{n = 0}^{\infty} \frac{1}{7} ^ n = {\sum}_{n = 0}^{\infty} {\left(\frac{1}{7}\right)}^{n} = \frac{1}{1 - \frac{1}{7}} = \frac{7}{6}$

${\sum}_{n = 0}^{8} \frac{1}{7} ^ n = {\sum}_{n = 0}^{8} {\left(\frac{1}{7}\right)}^{n} = {\left(1 - \frac{1}{7}\right)}^{9} / \left(1 - \frac{1}{7}\right) = \frac{7}{6} \left(1 - \frac{1}{40353607}\right) = \frac{282475242}{242121642}$

So:

$1 + \frac{1}{5} + \frac{1}{7} ^ 2 + \frac{1}{7} ^ 3 + \frac{1}{7} ^ 9 + \ldots + \frac{1}{7} ^ n + \ldots = 1 + \frac{1}{5} + \frac{1}{49} + \frac{1}{343} + \frac{7}{6} - \frac{282475242}{242121642} \cong 1.223323644$