# Math Help - Sum 3-1+1/3-1/9+...=9/4

1. ## Sum 3-1+1/3-1/9+...=9/4

I have numerically verified that the following sum comes to $\frac 94$, the same thing the back of the book says. So what'd I do wrong?

Find the sum of the convergent series
$3-1+\frac 13 - \frac 19 + \mathellipsis$.

I verified that the above sum can be represented as
$\sum_{n \ge 1} \frac{(-1)^{n+1}9}{3^n}$,

which can be rewritten as
$\sum_{n \ge 1} \frac{(-1)^{n+1}9}{3^n} = -9\sum_{n \ge 1} \frac{(-1)^{n}}{3^n} = -9\sum_{n \ge 1} \left(\frac{-1}{3}\right)^n$.

I believe I can apply the geometric series to the summation, giving me
$-9\sum_{n \ge 1} \left(\frac{-1}{3}\right)^n = -9 \cdot \frac{1}{1+\frac 13} = -9 \cdot \frac 34 = -\frac{27}{4} \ne \frac 94$.

Ahhh... but now I see that the geometric series equation starts the summation at $n=0$ (do you find that typing up your answers sometimes helps you realize what's wrong? Why is that?).

So, I need to subtract $-9 \cdot \frac{-1}{3}^0 = -9$ from my answer, which (unsurprisingly) gives $\frac 94$.

2. ## 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

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

Hello, MSUMathStdnt!

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

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

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

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