cherrilyn
  • cherrilyn
approximate the value of the series to within an error of at most 10^-5 ?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
cherrilyn
  • cherrilyn
\[\sum_{n=1}^{\infty} (1)^{n+1}\ln n/n! \]
cherrilyn
  • cherrilyn
I know |Sn-S|\[\le\] An+1 = ln(N+1)/(N+1)! \[\le 10^{-5}\]
watchmath
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

nowhereman
  • nowhereman
You can take N so that \[\frac{\ln(N+1)}{(N+1)!} \leq \frac{N+1}{(N+1)!} = \frac{1}{N!}\leq 10^{-5}\]
cherrilyn
  • cherrilyn
okayy. now how would I go about finding the exact answer?
nowhereman
  • nowhereman
Just add all the summands from N=1 to \(N! \leq 10^5\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.