Find all values of x for which the series converges. For these values of x,...

Consider the series
$\sum _{n=1}^{\mathrm{\infty }}(3x{)}^n$
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
$S_{\mathrm{\infty }}=\frac{a}{1-r}$
$\sum _{n=1}^{\mathrm{\infty }}(3x{)}^n=(3x{)}^1+(3x{)}^2+(3x{)}^3+(3x{)}^4....$
$\sum _{n=1}^{\mathrm{\infty }}(3x{)}^n=(3x)+(3x{)}^2+(3x{)}^3+(3x{)}^4....$
The given series is a Geometric progression with the
first term $=3x$
common ratio $=3x$
The given series is convergent for

$|3x|<1$

$\Rightarrow -1<3x<1$

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

Sum of the series

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

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

Here

$a=3x,r=3x$

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

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

Answer:

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

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

Answer is given below (on video)