Home » Blog » Prove that the sum from 1 to ∞ of (2n + 3n) / 6n = 3/2

Prove that the sum from 1 to ∞ of (2n + 3n) / 6n = 3/2

Prove that the following sum converges and has the given value.

    \[ \sum_{n=1}^{\infty} \frac{2^n + 3^n}{6^n} = \frac{3}{2}. \]


We have,

    \begin{align*}  \sum_{n=1}^{\infty} \frac{2^n + 3^n}{6^n} &= \sum_{n=1}^{\infty} \left( \left( \frac{1}{3} \right)^n + \left( \frac{1}{2} \right)^n \right) \\[9pt]  &= \sum_{n=1}^{\infty} \left( \frac{1}{3} \right)^n + \sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^n. \end{align*}

This last step is justified since each of the two series are geometric series; hence, the series of their sum converges and is equal to the sum of the two series by Theorem 10.2 on page 385 of Apostol. Then, using the formula for the sum of a geometric series we have

    \begin{align*}  \sum_{n=1}^{\infty} \frac{2^n  + 3^n}{6^n} &= \sum_{n=1}^{\infty} \left( \frac{1}{3} \right)^n + \sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^n \\[9pt]  &= \sum_{n=0}^{\infty} \left( \frac{1}{3}\right)^n - 1 + \sum_{n=0}^{\infty} \left( \frac{1}{2} \right)^n - 1 \\[9pt]  &= \frac{1}{1 - \frac{1}{3}} - 1 + \frac{1}{1 - \frac{1}{2}} - 1 \\[9pt]  &= \frac{3}{2}. \end{align*}

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):