Q 53. - Calculus

Question

Determine whether the series $\sum _{k=2}^{\infty }{\left(\frac{-3}{5}\right)}^k$ converges or diverges. Give the sum of the convergent series.

Step-by-Step Solution

Answer

The series $\sum _{k=2}^{\infty }{\left(\frac{-3}{5}\right)}^k$ converges to $\frac{9}{40}$.

Step 1. Given information.

Given a series $\sum _{k=2}^{\infty }{\left(\frac{-3}{5}\right)}^k$.

Step 2. Find if the series converges or not.

The index starts with 2, rather than 0.

Note that the convergence of a series depends not upon the first few terms but only upon the tail of the series.

The standard form of geometric series is $\sum _{k=0}^{\infty }c{r}^k$ .

Here, the series $\sum _{k=2}^{\infty }{\left(\frac{-3}{5}\right)}^k$ has $c=\frac{9}{25}$ and $r=\frac{-3}{5}$.

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

Since $r=\frac{-3}{5}$, it follows that the series $\sum _{k=2}^{\infty }{\left(\frac{-3}{5}\right)}^k$ 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 $\sum _{k=2}^{\infty }{\left(\frac{-3}{5}\right)}^k$ converges to $\frac{{\displaystyle \frac{9}{25}}}{1-\left({\displaystyle \frac{-3}{5}}\right)}$, that is $\frac{9}{40}$.