Question

how do I find the sum of this series?\

http://i.imgur.com/Phxfx.png

Answer 1

looks like geometric progression.

Answer 2

\( 1/3 \sum (2x^2/3)^n\)

if x^2 < 3/2, the series is convergent, else it is divergent.

Answer 3

\[
g(v)=\sum _{n=1}^{\infty } v^n=-\frac{v}{v-1}, \quad |v| <1
\]
Your sum is
\[
\frac 1 3 g\left( \frac{2 x^2}{3} \right)=-\frac{2 x^2}{2 x^2-3}
\]

Answer 4

In the above, we must have
\[ \frac{2 x^2}{3} <1

\]