# Simple Geometric Series

• Nov 21st 2009, 09:41 AM
WhoCares357
Simple Geometric Series
$\sum_{k=1}^\infty (\frac{2}{3})^{k+2}$

This is what I did.
$\sum_{k=1}^\infty \frac{4}{9}*(\frac{2}{3})^{k}$
$a=\frac{4}{9}, r=\frac{2}{3}<1$
Threfore series converges to
$\frac{4}{3}$

This seems correct to me. However when I try to solve it thorugh maple (software) it outputs the result of $\frac89$. Did I do something wrong?
• Nov 21st 2009, 09:58 AM
e^(i*pi)
Quote:

Originally Posted by WhoCares357
$\sum_{k=1}^\infty (\frac{2}{3})^{k+2}$

This is what I did.
$\sum_{k=1}^\infty \frac{4}{9}*(\frac{2}{3})^{k}$
$a=\frac{4}{9}, r=\frac{2}{3}<1$
Threfore series converges to
$\frac{4}{3}$

This seems correct to me. However when I try to solve it thorugh maple (software) it outputs the result of $\frac89$. Did I do something wrong?

Your value of $a$ is incorrect. As the sum starts at $k=1$ you'd need to do

$\frac{4}{9} \times \left(\frac{2}{3}\right)^1 = \frac{8}{27}$

Putting in this value of a should give $\frac{8}{9}$
• Nov 21st 2009, 10:00 AM
WhoCares357
Quote:

Originally Posted by e^(i*pi)
Your value of $a$ is incorrect. As the sum starts at $k=1$ you'd need to do

$\frac{4}{9} \times \left(\frac{2}{3}\right)^1 = \frac{8}{27}$

Putting in this value of a should give $\frac{8}{9}$

Can you explain why? Wouldn't that force the series to start from 2?
• Nov 21st 2009, 10:06 AM
e^(i*pi)
Quote:

Originally Posted by WhoCares357
Can you explain why? Wouldn't that force the series to start from 2?

As I understood it as it's the first term a would be the value of f(k) at k=1 as that's where the sum starts from.

I don't think the series starts at 2 as a is the initial value before anything is added
• Nov 21st 2009, 01:28 PM
RockHard
Just a question I am new to series, because this geometric series starts at n=1 would you not try to get the power in terms of n - 1?