# Trouble calculating an alternating/infinite sum

• Sep 14th 2011, 01:42 PM
s3a
Trouble calculating an alternating/infinite sum
Firstly, the sum is attached as a png file and the answer is that it converges (conditionally) to 1/12.

I thought about the fact that I could split that up to two sums yielding one with all the positive terms and one with all the negative terms and then evaluate the sums separately and then have simple arithmetic at the end but I don't know how to set up the positive/negative separation. In other words, I don't know what needs to be done mechanically in order to transform that one infinite sum into two infinite sums (one being positive and the other negative).

Any help would be greatly appreciated!

Edit: If I am completely off in my reasoning, just please tell me how to do it in the first place.
• Sep 14th 2011, 02:13 PM
skeeter
Re: Trouble calculating an alternating/infinite sum
Quote:

Originally Posted by s3a
Firstly, the sum is attached as a png file and the answer is that it converges (conditionally) to 1/12.

I thought about the fact that I could split that up to two sums yielding one with all the positive terms and one with all the negative terms and then evaluate the sums separately and then have simple arithmetic at the end but I don't know how to set up the positive/negative separation. In other words, I don't know what needs to be done mechanically in order to transform that one infinite sum into two infinite sums (one being positive and the other negative).

Any help would be greatly appreciated!
$\displaystyle \sum_{n=1}^{\infty} \frac{(-4)^{n-1}}{8^n} = -\frac{1}{4} \cdot \sum_{n=1}^{\infty} \frac{(-4)^{n}}{8^n} = -\frac{1}{4} \cdot \sum_{n=1}^{\infty} \left(-\frac{1}{2}\right)^n$
$\displaystyle S = -\frac{1}{4} \cdot \frac{-\frac{1}{2}}{1 - \left(-\frac{1}{2}\right)}$