Answer to Question #189971 in Calculus for Moel Tariburu

Question

Find the sum of the infinite series ∑_(n=1)^(+∞)▒2^(k+1)/3^(k+2) 

Answer 1

First let us rewrite it as

$\sum_{k\geq 1} \frac{2^{k+1}}{3^{k+2}}=\frac{2}{3^2}\sum_{k\geq 1} \frac{2^k}{3^k} = \frac{2}{9} \sum_{k\geq 1} (\frac{2}{3})^k$

The last series is a geometric series, for which we have

$\sum_{k\geq 1} q^k = \frac{q}{1-q}$ (which we can deduce from $\sum_{k=1}^n q^k = \frac{q-q^{n+1}}{1-q}$ and $q^{n+1}\to 0$ for $|q|<1$) and therefore the result is given by

$\frac{2}{9} \sum_{k\geq 1} (\frac{2}{3})^k =\frac{2}{9} \times \frac{\frac{2}{3}}{1-\frac{2}{3}}=\frac{4}{9}$