Q 58. - Calculus

Question

Determine whether the series $\sum _{k=0}^{\infty }\frac{5^{k+1}}{{\left(-6\right)}^k}$ converges or diverges. Give the sum of the convergent series.

Step-by-Step Solution

Answer

The series $\sum _{k=0}^{\infty }\frac{5^{k+1}}{{\left(-6\right)}^k}$ converges to $\frac{30}{11}$.

Step 1. Given information.

Given a series $\sum _{k=0}^{\infty }\frac{5^{k+1}}{{\left(-6\right)}^k}$.

Step 2. Find if the series converges or not.

The series $\sum _{k=0}^{\infty }\frac{5^{k+1}}{{\left(-6\right)}^k}$ can be expressed as $\underset{k=0}{\overset{\infty }{\sum 5}}{\left(\frac{-5}{6}\right)}^k$.

The series $\sum _{k=0}^{\infty }\frac{5^{k+1}}{{\left(-6\right)}^k}$ is in the standard form $\sum _{k=0}^{\infty }c{r}^k$ for a geometric series with $c=5$ and $r=-\frac{5}{6}$.

The geometric series converges if and only if $\left|r\right|<1$.

Since $r=-\frac{5}{6}$, it follows that the series converges.

Step 3. Find the value to which the series converges.

If the geometric series $\sum _{k=0}^{\infty }c{r}^k$ converges, it converges to $\frac{c}{1-r}$.

So, the series $\sum _{k=0}^{\infty }\frac{5^{k+1}}{{\left(-6\right)}^k}$ converges to $\frac{5}{1-\left(-{\displaystyle \frac{5}{6}}\right)}$, that is $\frac{30}{11}$.