## cherrilyn 5 years ago approximate the value of the series to within an error of at most 10^-5 ?

1. cherrilyn

$\sum_{n=1}^{\infty} (1)^{n+1}\ln n/n!$

2. cherrilyn

I know |Sn-S|$\le$ An+1 = ln(N+1)/(N+1)! $\le 10^{-5}$

3. watchmath

Then from there we just make a guess what is the smallest n that will do the job. So you pick some n if it doesn't work, try n+1 an so on.

4. nowhereman

You can take N so that $\frac{\ln(N+1)}{(N+1)!} \leq \frac{N+1}{(N+1)!} = \frac{1}{N!}\leq 10^{-5}$

5. cherrilyn

Just add all the summands from N=1 to $$N! \leq 10^5$$