# Geometric Series Problem

• Apr 27th 2009, 10:46 AM
eg37se
Geometric Series Problem
Hey guys, I am currently in a DiffEq class, and I can't remember how to do Geometric Series for the life of me. I appreciate any help you can give me.

∑[(e^(-(i+1)πs)+e^(-iπs))/(s^2+1)]

The summation is from 0 to infinity.
• Apr 27th 2009, 10:50 AM
Jester
Quote:

Originally Posted by eg37se
Hey guys, I am currently in a DiffEq class, and I can't remember how to do Geometric Series for the life of me. I appreciate any help you can give me.

∑[(e^(-(i+1)πs)+e^(-iπs))/(s^2+1)]

The summation is from 0 to infinity.

What's the index n or s?
• Apr 27th 2009, 11:02 AM
eg37se
Ohhh yea sorry, it looks confusing.

∑[(e^(-(i+1)πs)+e^(-iπs))/(s^2+1)]

s is a constant
i is the index
π=pi
• Apr 27th 2009, 11:33 AM
Jester
Your series is $\frac{1}{1+s^2} \sum_{i=0}^\infty e^{-(i+1) \pi s} + e^{- i \pi s}$ which can be written as

$
\frac{1 + e^{\pi s}}{1+ s^2} \sum_{i=0}^\infty e^{- i \pi s} =
\frac{1 + e^{\pi s}}{1+ s^2} \sum_{i=0}^\infty \left(e^{- \pi s} \right)^i
$
= $
\frac{1 + e^{\pi s}}{1+ s^2} \cdot \frac{1}{1 - e^{-\pi s}}
$
provided that $e^{-\pi s} < 1$.

In general for infinite geometric series $S = a + ar + ar^2 + ar^3 + \cdots = \frac{a}{1-r}$ provided that $|r| < 1.$
• Apr 27th 2009, 09:40 PM
eg37se
Thanks a lot, I appreciate it.