Expert: Sherry Wallin - 6/1/2009

Question
Hey,
I have another question for you if you get a chance.
Approximate the sum of the series Sigma(with an infinity sign above it and n=0 under it) of [(-3)^n]/[(2^n)n!]
Sorry, I am not so great at setting up the question, I don't really know how to write out sigma notation but I did it in an elementary fashion.
Hope you can help,
 All the best, Janine

Answer
Janine~
    You can look at the series as being the sum of [(-3/2)^n]/n! Note that n! grows faster than 2^n for n > 3 so the sum of [(3/-2)^n]/n!<
sum of[(3/-2)^n]/2^n = (-3)^n/2^(2n)which goes quite rapidly to 0 as n gets large. Hence your original sum will move even quicker to 0 as n gets large. Does this make sense?

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