Summation of: $\sum_{1}^{\infty}\left(\frac{2}{3}\right)^x$

Question

This is a subsection in my statistics homework. It goes back to calculus II and summations, and it's been a long time since I've studied it so I'm rusty.

I'm looking to solve the summation of $c\left(\frac{2}{3}\right)^x$ where $c$ is a constant to where I can solve for $c$.

I understand that we say that this summation is: $$c\left[\frac{2}{3}+\left(\frac{2}{3}\right)^2 +\left(\frac{2}{3}\right)^3 +\cdots\right]$$ But the book then says that this then equals to:

$$c\dfrac{\dfrac{2}{3}}{1-\dfrac{2}{3}}$$

Any help?

Also I know my coding ability is little to none so if you'd like to edit my question to make it look more presentable feel free to.

Thanks all.

Answer 1

This is a geometric series, and it converges since $2/3 <1$.

If we write $$S=c ( (2/3) + (2/3)^2 + (2/3)^3 + \cdots )$$ then we see that $$(2/3) S = c ( (2/3)^2 + (2/3)^3 + (2/3)^4 + \cdots )$$

subtracting $(2/3)S$ from $S$ we find: $$( 1-(2/3) )S = c ( 2/3)$$ Now if we solve for $S$ we have: $$S = \frac{c(2/3)}{1-(2/3)}$$ which is the answer provided in your homework.