Calcular el $\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}$

Question

Calcular el $\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}$

\begin{align}\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}&=3\sum_{k=2}^{\infty}\frac{1}{5^{k-1}}\\&=3\sum_{k=2}^{\infty}\frac{1}{5^{k}}\cdot\frac{1}{5^{-1}}\\&=3\sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k}\cdot5\\&= 15\left( \sum_{k=0}^{\infty}\left( \frac{1}{5} \right )^{k}-2 \right )\\&=15\left( \frac{1}{1-\frac{1}{5}}-2 \right )\\&=-\frac{45}{4}\end{align}

Hice correctamente?

Answer 1

Este paso es malo: $3\sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k}\cdot5= 15\left( \sum_{k=0}^{\infty}\left( \frac{1}{5} \right )^{k}-2 \right )$

Debe ser: $3\sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k}\cdot5= 15\left( \sum_{k=2}^{\infty}\left( \frac{1}{5} \right )^{k} \right)=15(\sum_{k=0}^\infty\frac{1}{5}^k-\frac{6}{5})=15(\frac{1}{1-\frac{1}{5}}-\frac{6}{5})=15(\frac{5}{4}-\frac{6}{5})=15\frac{1}{20}=\frac{3}{4}$

Answer 2

Usted shoould obtener un número positivo. Tal vez usted podría seguir este enfoque más sencillo: $$\sum_{k=2}^{\infty}\frac{3}{5^{k-1}}=\frac{3}{5}\sum_{k=2}^{\infty}\frac{1}{5^{k-2}}=\frac{3}{5}\sum_{j=0}^{\infty}\frac{1}{5^{j}}=\frac{3}{5}\cdot\frac{1}{1-1/5}=\frac{3}{4}.$$

Answer 3

Su suma debe ser $$15\times \sum \limits^{\infty }_{k=2}\left( \frac{1}{5} \right) ^{k}=15\times \frac{\left( \frac{1}{5} \right) }{1-\frac{1}{5} } ^{2}=15\times \frac{1}{20} =\frac{3}{4} $$