Use the formula ln(1+x)= summation from n=0 to infinity of ((-1)^(n-1)(x^n))/(n)

Question

Use the formula

ln(1+x)= summation from n=0 to infinity of ((-1)^(n-1)(x^n))/(n) = x - ((x^2)/2) + ((x^3)/3) - ((x^4)/4) + .......... for |x|<1

and your knowledge of alternating series to find the smallest N such that the partial sum S_N approximates ln(9/8) to within an error of at most 10^-3. Confirm this by computing both S_N and ln(9/8).  (round answers to six places)

N=__________________

S_N=________________

ln(9/8)=______________

Error = absolute value of ln(9/8)-S_N) =____________________

Explanation / Answer

We know that for this alernating series

|S - S_N| <= |a_(N+1)|

where a_(N+1) is the (N+1)st term

So we want to find N such that

|a_(N+1)| < .001

In our case, x = 1/8

(1/8)^(N+1) / (N+1) < .001

Note that (1/8)^2/2 = 0.0078125 > .001

But (1/8)^3 / 3 = 0.00065104166 < .001

So N+1 = 3

N = 2

We just need the first two terms

ln( 9/8) = ln (1 + (1/8)) is approximately equal to

S_2 = 1/8 - (1/8)^2 / 2 = .125 - 0.0078125 = 0.117188 (rounded to six decimal places)

The true value of ln (9/8) is 0.117783 (rounded to six decimal places)

So our estimate is within .001 of the true value.

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