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

by RoRi

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*}

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