(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Ʃ(1 to infinity) (2/3)^(3n)

2. Relevant equations

For a geometric series, the series converges to a/1-r

3. The attempt at a solution

I'm really just confused about how to manipulate this so that it has a form of ar^n, especially since it starts at 1 rather than 0. I know that since the series starts at 1, we ought to have it in the form ar^n-1, but I'm at a loss as to how to go about.

I'm not really concerned with the answer so much as how one would go about getting it, and really understanding the process. It seems I'm having a hard time figuring out how to manipulate series and get them in the correct forms...

Any help in the right direction would be great. Thanks!

This is what I've come up with, but I'm not sure if it's correct...

We can rewrite the series as ((2/3)^3)^k = (8/27)^(k-1) * (8/27)

a = 8/27

r = 8/27

Computing a/1-r, I get 8/19...

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# Homework Help: Find the Sum of The Geometric Series

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