?How do you find the sum of the infinite geometric series 2^n/5^(2n+1)?

Nov 27, 2015

${\sum}_{n = 0}^{\infty} {2}^{n} / {5}^{2 n + 1} = \frac{5}{23}$

Explanation:

The sum of an infinite geometric series with ratio $r$ where $| r | < 1$ is given by
${\sum}_{n = 0}^{\infty} a {r}^{n} = \frac{a}{1 - r}$

(A quick derivation for this formula is included in this answer: Can an infinite series have a sum? )

Now
${2}^{n} / {5}^{2 n + 1} = \left(\frac{1}{5}\right) \left({2}^{n} / {5}^{2 n}\right) = \left(\frac{1}{5}\right) \left({2}^{n} / {25}^{n}\right) = \left(\frac{1}{5}\right) {\left(\frac{2}{25}\right)}^{n}$

So, as $| \frac{2}{25} | < 1$

we have
${\sum}_{n = 0}^{\infty} {2}^{n} / {5}^{2 n + 1} = {\sum}_{n = 0}^{\infty} \left(\frac{1}{5}\right) {\left(\frac{2}{25}\right)}^{n} = \frac{\frac{1}{5}}{1 - \left(\frac{2}{25}\right)} = \frac{5}{23}$