 ka1leE

2020-10-20

Show that the series converges. What is the value of the series?
$\sum _{n=2}^{\mathrm{\infty }}\left(-\frac{5}{3}{\right)}^{n}\left(\frac{2}{5}{\right)}^{n+1}$ We have to Show that the series converges. What is the value of the series?
Series is given below:
$⇒\sum _{n=2}^{\mathrm{\infty }}\left(-\frac{5}{3}{\right)}^{n}\left(\frac{2}{5}{\right)}^{n+1}$
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:
$⇒\sum _{n=1}^{\mathrm{\infty }}a{r}^{n-1}$
$|r|<1$
$⇒{S}_{\mathrm{\infty }}=\frac{a}{1-r}$
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:
$⇒\sum _{n=2}^{\mathrm{\infty }}\left(-\frac{5}{3}{\right)}^{n}\left(\frac{2}{5}{\right)}^{n+1}$
$⇒\sum _{n=2}^{\mathrm{\infty }}\left(-\frac{5}{3}{\right)}^{n}\left(\frac{2}{5}{\right)}^{n+1}$
$⇒\sum _{n=2}^{\mathrm{\infty }}\frac{2}{5}\left(-1{\right)}^{n}\left(\frac{2}{3}{\right)}^{n}$
$⇒|r|=|-\frac{2}{3}|$
$|r|=\frac{2}{3}<1$
Previous step continue
${⇒}_{n=2}^{\mathrm{\infty }}\frac{2}{5}\left(-1{\right)}^{n}\left(\frac{2}{3}{\right)}^{n}$
$⇒\frac{8}{45}-\frac{16}{135}+\frac{32}{405}...$
$⇒a=\frac{8}{45}$
$⇒r=-\frac{2}{3}$
$⇒{S}_{\mathrm{\infty }}=\frac{\frac{8}{45}}{1+\frac{2}{3}}$
$=\frac{8}{75}$
$⇒{S}_{\mathrm{\infty }}=\frac{8}{75}$ Jeffrey Jordon