# Alternating Series Estimation Theorem

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1. May 6, 2017

### jlmccart03

1. The problem statement, all variables and given/known data
Using the power series for ln(x + 1) and the Estimation Theorem for the Alternating Series, we conclude that the least number of terms in the series needed to approximate ln 2 with error < 3/1000 is: (i) 333 (ii) 534 (iii) 100 (iv) 9 (v) 201

2. Relevant equations
ln(x+1) = Σ(-1)^nx^n/n!

3. The attempt at a solution
I know that the alternating series estimation thm is |Sn-S| ≤ (estimation) which is 3/1000. I get x to be equal to 1 since we want ln(2), but when I setup the equation I get lost on how to simplify this to a specific n value. (Calculators are not allowed on exam so I am rusty with algebra).

I get (-1)^n+1 * 2^n/(n+1)! ≤ 3/1000 which gives 2^n ≤ 3/1000 * (n+1) and I can't figure how to get the n in the exponent down without using ln yet the answers are specific numbers.

2. May 6, 2017

### ehild

For an alternating series, the estimated error is less than the magnitude of the term following the last one.
What is the Taylor series of ln(x+1) around x=0?

3. May 6, 2017

### Ray Vickson

(1) You have the wrong series; your series is the expansion of $e^{-x}$, which is not what was asked. The expansion of $\ln(1+x)$ is a lot simpler, and produces a problem solvable easily in an exam setting without a calculator.
(2) You say you need $x = 1$, so how did $1^n$ become $2^n$?