Geometric Series Problem

**eg37se** · April 27th 2009, 10:46 AM

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.

**Danny** · April 27th 2009, 10:50 AM

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?

**eg37se** · April 27th 2009, 11:02 AM

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

**Danny** · April 27th 2009, 11:33 AM

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 $|r| < 1.$

**eg37se** · April 27th 2009, 09:40 PM

Thanks a lot, I appreciate it.

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