Alternating Series Estimation Theorem

[jlmccart03]

Homework Statement

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

Homework Equations

ln(x+1) = Σ(-1)^nx^n/n!

The Attempt at a Solution

I know that the alternating series estimation thm is |S_n-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.

 

[ehild]

Homework Statement

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

Homework Equations

ln(x+1) = Σ(-1)^nx^n/n!

The Attempt at a Solution

I know that the alternating series estimation thm is |S_n-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.

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?

 

[Ray Vickson]

Homework Statement

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

Homework Equations

ln(x+1) = Σ(-1)^nx^n/n!

The Attempt at a Solution

I know that the alternating series estimation thm is |S_n-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.

(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 $2^n$?