Re: Sum 3-1+1/3-1/9+...=9/4

Hey MSUMathStdnt.

Given the convergence criteria the sum should be 3 - (1/[1+1/3]) = 3 - 3/4 = 12/4 - 3/4 = 9/4

Re: Sum 3-1+1/3-1/9+...=9/4

Hello, MSUMathStdnt!

Quote:

$\displaystyle \text{Find the sum: }\:3-1+\frac 13 - \frac 19 + \hdots$

This is a geometric series with first term $\displaystyle a = 3$ and common ratio $\displaystyle r = \text{-}\tfrac{1}{3}$

The sum is: .$\displaystyle S \:=\:\frac{a}{1-r}$

Therefore: .$\displaystyle S \;=\;\dfrac{3}{1-\left(\text{-}\frac{1}{3}\right)} \;=\;\frac{3}{\frac{4}{3}} \;=\;\frac{9}{4}$