Question

Calculate E:

Answer 1

The key thing here is to group everything under the same exponent

Answer 2

if i grouped everything correctly,

Answer 3

\[E=\sum\limits_{n=1}^{\infty} (-2)^n\cdot \dfrac{2}{3}\cdot \dfrac{1}{3^{n-1}} = 2\sum\limits_{n=1}^{\infty} (-2)^n\cdot \dfrac{1}{3^{n}} =2\sum\limits_{n=1}^{\infty} \left(-\dfrac{2}{3}\right)^n \]

Answer 4

that looks like the familiar geometric series ?

Answer 5

yes, do i have to find the limit then?