We have to Show that the series converges. What is the value of the series?

Series is given below:

\(\Rightarrow\sum_{n=2}^\infty(-\frac{5}{3})^n(\frac{2}{5})^{n+1}\)

We will use geometric series test .

Geometriic series test says if series of the form \(ar^{n-1}\) and \(|r|<1\) then we can say that series is converges and sum of series is given by S.

Work is shown below:

\(\Rightarrow\sum_{n=1}^\infty ar^{n-1}\)

\(|r|<1\)

\(\Rightarrow S_{\infty}=\frac{a}{1-r}\)

Now compare the given series to geometric series and find value of r .

If \(r<1\) then series converges.

In this case \(r=\frac23 <1\) so series converge.

After that find value of a (starting term of series).

With the help of a, r and sum of geometric series formula we will find sum.

Work is shown below:

\(\Rightarrow\sum_{n=2}^\infty(-\frac53)^n(\frac25)^{n+1}\)

\(\Rightarrow\sum_{n=2}^\infty(-\frac53)^n(\frac25)^{n+1}\)

\(\Rightarrow\sum_{n=2}^\infty\frac25(-1)^n(\frac23)^n\)

\(\Rightarrow|r|=|-\frac23|\)

\(|r|=\frac23<1\)

Previous step continue

\(\Rightarrow_{n=2}^\infty\frac25(-1)^n(\frac23)^n\)

\(\Rightarrow\frac{8}{45}-\frac{16}{135}+\frac{32}{405}...\)

\(\Rightarrow a=\frac{8}{45}\)

\(\Rightarrow r=-\frac{2}{3}\)

\(\Rightarrow S_{\infty}=\frac{\frac{8}{45}}{1+\frac23}\)

\(=\frac{8}{75}\)

Answer

\(\Rightarrow S_{\infty}=\frac{8}{75}\)