Show that the series converges. What is the value of the series? sum_{n=2}^infty(-frac{5}{3})^n(frac{2}{5})^{n+1}

We have to Show that the series converges. What is the value of the series?
Series is given below:
$\Rightarrow \sum _{n=2}^{\mathrm{\infty }}(-\frac{5}{3}{)}^n(\frac{2}{5}{)}^{n+1}$
We will use geometric series test .
Geometriic series test says if series of the form $a{r}^{n-1}$ and $|r|<1$ then we can say that series is converges and sum of series is given by S.
Work is shown below:
$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}a{r}^{n-1}$
$|r|<1$
$\Rightarrow S_{\mathrm{\infty }}=\frac{a}{1-r}$
Now compare the given series to geometric series and find value of r .
If $r<1$ then series converges.
In this case $r=\frac{2}{3}<1$ 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:
$\Rightarrow \sum _{n=2}^{\mathrm{\infty }}(-\frac{5}{3}{)}^n(\frac{2}{5}{)}^{n+1}$
$\Rightarrow \sum _{n=2}^{\mathrm{\infty }}(-\frac{5}{3}{)}^n(\frac{2}{5}{)}^{n+1}$
$\Rightarrow \sum _{n=2}^{\mathrm{\infty }}\frac{2}{5}(-1{)}^n(\frac{2}{3}{)}^n$
$\Rightarrow |r|=|-\frac{2}{3}|$
$|r|=\frac{2}{3}<1$
Previous step continue
${\Rightarrow }_{n=2}^{\mathrm{\infty }}\frac{2}{5}(-1{)}^n(\frac{2}{3}{)}^n$
$\Rightarrow \frac{8}{45}-\frac{16}{135}+\frac{32}{405}...$
$\Rightarrow a=\frac{8}{45}$
$\Rightarrow r=-\frac{2}{3}$
$\Rightarrow S_{\mathrm{\infty }}=\frac{\frac{8}{45}}{1+\frac{2}{3}}$
$=\frac{8}{75}$
Answer
$\Rightarrow S_{\mathrm{\infty }}=\frac{8}{75}$