1. ## 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?

2. 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 $\frac{8}{9}$

3. 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?

4. 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

5. 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?