[HARD] Sigma (n=0 to infinity) ((-1)^n)/(9^n)n! (see below)
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[HARD] Sigma (n=0 to infinity) ((-1)^n)/(9^n)n! (see below)

[From: ] [author: ] [Date: 12-04-25] [Hit: ]
..Note that e^x= 1 + x + x^2/2! + x^3/3! + x^4/4!.......
Wrong on test. Help? 1. Identify Bn. 2. Find limit of Bn 3. If the series is convergent, use the Alternating Series Estimation Theorem to determine how many terms we need to add in order to find the sum with an error less than 0.000005.

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-(1/9)/1! + (1/9)^2/2! - (1/9)^3/3! + (1/9)^4/4! ... = sum
Note that e^x= 1 + x + x^2/2! + x^3/3! + x^4/4!... and is absolutely convergent for EVERY x.
So e^(-1/9) = 1 - (1/9)/1! + (1/9)^2/2! - (1/9)^3/3! + (1/9)^4/4! ...
So e^(-1/9) = 1+sum
sum=e^(-1/9) - 1

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We have the well known series
exp(-x)= 1- x +x^2/2! -x^3/3! + x^4/4! - .....
Putting x=1/9, we find the series sums to exp(-1/9).
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