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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.

Michael Hardy
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Cole Howie
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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.

Joel
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  • is there any set rule you can give me for future problems like this? I've tried to google around but I didn't find any that gave an answer of r/1-r such as my homework says the answer is – Cole Howie Sep 08 '14 at 14:46
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    @ColeHowie I promise you it is there in the wiki. – David H Sep 08 '14 at 14:53
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    @ColeHowie : You might Google the term "geometric series". A geometric series is a series with a common ratio, i.e. the number you multiply a term by to get the next term is the same in all cases. ${}\qquad{}$ – Michael Hardy Sep 08 '14 at 14:54