Sorry about the title, if possible please change it

[tex]\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} + ...) [/tex]

Using the geometric series,

[tex]\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} + ...) [/tex]

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

substituting into the original problem:

[tex]\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} + ... ) [/tex]

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

**1. Find the sum of the following convergent series**

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

**2. [tex]\sum_{j=0}^{\infty}c^{j} = 1/(1-c) if |c| < 1[/tex]****3. The Attempt at a Solution**[tex]\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} + ...) [/tex]

Using the geometric series,

[tex]\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} + ...) [/tex]

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

substituting into the original problem:

[tex]\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} + ... ) [/tex]

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

Last edited: