infinite series - 11

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \sum_{k=0}^{\infty}\frac{1}{2}\left(-\frac{1}{3}\right)^k\)

 

[logistic_guy]

Let us pretend that this is a geometric series.

Then,

\(\displaystyle \sum_{k=0}^{\infty}\frac{1}{2}\left(-\frac{1}{3}\right)^k = \frac{1/2}{1 - \left(-\frac{1}{3}\right)} = \textcolor{indigo}{\frac{3}{8}}\)