beljuA

2021-02-25

Find all values of x for which the series converges. For these values of x, write the sum of the series as a function of x.
$\sum _{n=1}^{\mathrm{\infty }}\left(3x{\right)}^{n}$

tabuordg

Consider the series
$\sum _{n=1}^{\mathrm{\infty }}\left(3x{\right)}^{n}$
Consider a G.P series $\sum _{n=1}^{\mathrm{\infty }}{r}^{n}$
Series is convergent for $|r|<1$
Sum of this infinite series is given by
${S}_{\mathrm{\infty }}=\frac{a}{1-r}$
$\sum _{n=1}^{\mathrm{\infty }}\left(3x{\right)}^{n}=\left(3x{\right)}^{1}+\left(3x{\right)}^{2}+\left(3x{\right)}^{3}+\left(3x{\right)}^{4}....$
$\sum _{n=1}^{\mathrm{\infty }}\left(3x{\right)}^{n}=\left(3x\right)+\left(3x{\right)}^{2}+\left(3x{\right)}^{3}+\left(3x{\right)}^{4}....$
The given series is a Geometric progression with the
first term $=3x$
common ratio $=3x$
The given series is convergent for

$|3x|<1$

$⇒-1<3x<1$

$⇒-\frac{1}{3}$

Sum of the series

$\sum _{n=1}^{\mathrm{\infty }}\left(3x{\right)}^{n}$

for $-\frac{1}{3}$

Here

$a=3x,r=3x$

${S}_{\mathrm{\infty }}=\frac{a}{1-r}$

${S}_{\mathrm{\infty }}=\frac{3x}{1-3x}$

Series is convergent for $-\frac{1}{3}$

Sum of the series is $\frac{3x}{1-3x}$

Jeffrey Jordon