1. ## 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.

2. 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?

3. 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

4. Your series is $\displaystyle \frac{1}{1+s^2} \sum_{i=0}^\infty e^{-(i+1) \pi s} + e^{- i \pi s}$ which can be written as

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

In general for infinite geometric series $\displaystyle S = a + ar + ar^2 + ar^3 + \cdots = \frac{a}{1-r}$ provided that $\displaystyle |r| < 1.$

5. Thanks a lot, I appreciate it.