How do you find the sum of (1+1)+(1/3+1/5)+(1/9+1/25)+...+(3^-n+5^-n)+...?

Feb 13, 2017

${\sum}_{n = 0}^{\infty} \left({3}^{- n} + {5}^{- n}\right) = \frac{11}{4}$

Explanation:

We have:

${\sum}_{n = 0}^{\infty} \left({3}^{- n} + {5}^{- n}\right) = {\sum}_{n = 0}^{\infty} {\left(\frac{1}{3}\right)}^{n} + {\sum}_{n = 0}^{\infty} {\left(\frac{1}{5}\right)}^{n}$

Both these series are geometric series so the sum is:

${\sum}_{n = 0}^{\infty} \left({3}^{- n} + {5}^{- n}\right) = \frac{1}{1 - \frac{1}{3}} + \frac{1}{1 - \frac{1}{5}} = \frac{3}{2} + \frac{5}{4} = \frac{11}{4}$