Find the sum of the following convergent series

[PAR]

Sorry about the title, if possible please change it

1. Find the sum of the following convergent series

\sum_{j=0}^{\infty}(-1)^{j}(2/3)^{j}
2. \sum_{j=0}^{\infty}c^{j} = 1/(1-c) if |c| < 1

The Attempt at a Solution

\sum_{j=0}^{\infty}(-1)^{j}(2/3)^{j} = 1 - 2/3 + (2/3)^{2} + ...<br /> <br /> = 1 - (2/3 + (2/3)^{3} + ... ) + ((2/3)^{2} + (2/3)^{4} + ...)

Using the geometric series,
\sum_{j=0}^{\infty}(2/3)^{j} = 1 + (2/3) + (2/3)^{2} + ... = 1/(1-(2/3)) = 3<br /> = 1 + (2/3 + (2/3)^{3} + ...) + ((2/3)^{2} + (2/3)^{4} + ...)

(2/3)^{2} + (2/3)^{4} + ...) = 2 - (2/3 + (2/3)^{3} + ...)

substituting into the original problem:

\sum_{j=0}^{\infty}(-1)^{j}(2/3)^{j} = 1 - (2/3 + (2/3)^{3} + ... ) + (2 - (2/3 + (2/3)^{3} + ...) <br /> <br /> = 3 - 2(2/3 + (2/3)^{3} + ... )

Now i don't know what to do, would like some help, thanks!

 

[rochfor1]

Just plug write (-1)^j(2/3)^j as (-2/3)^j and use c=-2/3. c doesn't have to be positive, just of absolute value less than one.