Problem: What is the sum of the following series? \[ \left( \frac{1}{2\cdot 3} + \frac{1}{2^2 \cdot 3} \right) + \left( \frac{1}{2^2\cdot 3^2} + \frac{1}{2^3 \cdot 3^2} \right) + \cdots + \left( \frac{1}{2^n\cdot 3^n} + \frac{1}{2^{n+1} \cdot 3^n} \right) + \cdots \]
Solution: Note that we want to evaluate \begin{align*} \sum_{n=1}^\infty \left( \frac{1}{2^n\cdot 3^n} + \frac{1}{2^{n+1} \cdot 3^n} \right) & = \sum_{n=1} ^{\infty} \left( \frac{1}{2^n\cdot 3^n} + \frac{1}{2}\cdot\frac{1}{2^{n} \cdot 3^n} \right) \\[1ex] & = \sum_{n=1}^\infty \frac{1}{2^n\cdot 3^n}\left(1+\frac{1}{2}\right) \\[1ex] & = \frac{3}{2} \sum_{n=1}^{\infty} \frac{1}{2^n\cdot 3^n}. \end{align*} Now we recall that \[ \sum_{n=1}^\infty x^n = \frac{x}{1-x},~\iff |x| \lt 1. \] Here we have \[ \sum_{n=1} ^{\infty} \left( \frac{1}{6} \right) ^n. \] Since $\frac{1}{6} \lt 1$, so \[ \sum_{n=1} ^{\infty} \left( \frac{1}{6} \right) ^n = \frac{\frac{1}{6}}{1-\frac{1}{6}} = \frac{1}{5}. \] Therefore, \[ \sum_{n=1}^\infty \left( \frac{1}{2^n\cdot 3^n} + \frac{1}{2^{n+1} \cdot 3^n} \right) = \frac{3}{2}\sum_{n=1} ^{\infty} \left( \frac{1}{6} \right) ^n = \frac{3}{10}. \]