# Find the sum of the following convergent series

1. Mar 25, 2009

### PAR

1. Find the sum of the following convergent series

$$\sum_{j=0}^{\infty}(-1)^{j}(2/3)^{j}$$

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

3. The attempt at a solution

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

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

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

substituting into the original problem:

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

Now i dont know what to do, would like some help, thanks!

Last edited: Mar 25, 2009
2. Mar 25, 2009

### rochfor1

Re: rrr

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.