Question

Calculo III - Determine a soma da série

Answer 1

Olá!

Lembre da convergência da série geométrica:

$\displaystyle \sum_{n=0}^{\infty}r^n\to\dfrac{1}{1-r},\;\;\text{se}\;\;|r|<1.$

Temos:

$\displaystyle \sum_{n=0}^{\infty} \dfrac{3^n-2^n}{6^n}=\sum_{n=0}^{\infty}\dfrac{3^n}{6^n}-\sum_{n=0}^{\infty}\dfrac{2^n}{6^n} = \sum_{n=0}^{\infty}\left(\dfrac{1}{2}\right)^n-\sum_{n=0}^{\infty}\left(\dfrac{1}{3}\right)^n=\\ \\ \\ = \dfrac{1}{1-\frac{1}{2}}-\dfrac{1}{1-\frac{1}{3}}=\dfrac{1}{\frac{1}{2}}-\dfrac{1}{\frac{2}{3}}=2-\dfrac{3}{2}=\dfrac{1}{2}.$

Bons estudos!