Question

Calcule a soma do somatório de (1/2^n + 1/3^n) com n variando de 1 à ∞ ? "Caso tenham dúvida do enunciado vide anexo com a questão"

Answer 1

Boa noite Silva!

Solução!

$\displaystyle \sum_{n=1}^{\infty}\bigg( \frac{1}{2^{n} } + \frac{1}{3^{n}}\bigg)\\\\\\\ Reescrevendo~~o~~somatorio!\\\\\\\\ \displaystyle \sum_{n=1}^{\infty}\bigg( \frac{1}{2^{n} }\bigg) +\displaystyle\sum_{n=1} ^{\infty} \bigg(\frac{1}{3^{n}}\bigg)\\\\\\\ n=\{1,2,3,............\}\\\\\ a1= \dfrac{1}{2^{1} }= \dfrac{1}{2} \\\\\\\ a2= \dfrac{1}{2^{2} }= \dfrac{1}{4} \\\\\\\ a3= \dfrac{1}{2^{3} }= \dfrac{1}{8} \\\\\\\\\\\ S_{1} = \dfrac{a1}{1-q}$

 

$S_{1} = \dfrac{1}{ \dfrac{2}{ 1-\dfrac{1}{2} } }\\\\\\ S_{1} = \dfrac{1}{ \dfrac{2}{\dfrac{2-1}{2} } }\\\\\\ S_{1} = \dfrac{1}{ \dfrac{2}{\dfrac{1}{2} } }\\\\\\ S_{1} = \dfrac{1}{2 }\times\dfrac{2}{1} \\\\\\ \boxed{S_{1} = 1}$

Fazendo o mesmo processo para o segundo somatório.

$\displaystyle\sum_{n=1} ^{\infty} \bigg(\frac{1}{3^{n}}\bigg)\\\\\\\ n=\{1,2,3,............\}\\\\\\\\ a1= \dfrac{1}{3^{1} } = \dfrac{1}{3}\\\\\\\ a2= \dfrac{1}{3^{2} } = \dfrac{1}{9}\\\\\\\ a1= \dfrac{1}{3^{3} } = \dfrac{1}{27}$

$S_{2} = \dfrac{1}{ \dfrac{3}{ 1-\dfrac{1}{3} } }\\\\\\ S_{2} = \dfrac{1}{ \dfrac{3}{\dfrac{3-1}{3} } }\\\\\\ S_{2} = \dfrac{1}{ \dfrac{3}{\dfrac{2}{3} } }\\\\\\ S_{2} = \dfrac{1}{3 }\times\dfrac{3}{2} \\\\\\ \boxed{S_{2} = \frac{1}{2} }$

$S_{\infty}=S1+S2\\\\\\\ S_{\infty}=1+ \dfrac{1}{2}\\\\\\ S_{\infty}=\dfrac{2+1}{2}\\\\\\ S_{\infty}=\dfrac{3}{2}$

$\boxed{Resposta:~~\displaystyle \sum_{n=1}^{\infty}\bigg( \frac{1}{2^{n} } + \frac{1}{3^{n}}\bigg)= \frac{3}{2}}$

Boa noite!

Bons estudos!