anonymous
  • anonymous
need to find if converges or diverges using ratio test: sum of all (ln(n))/(n) when n=1 to inf *not sure what cancels once RT is set up
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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
any idea?
anonymous
  • anonymous
...why the ratio test? painful.
anonymous
  • anonymous
The ratio I get using the ratio test is 1. It wouldn't be a conclusive test.

Looking for something else?

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

More answers

anonymous
  • anonymous
If you want me to show you that, let me know.
anonymous
  • anonymous
it's not that painful loki all you've got to do is put : an+1 / an , I find it rather simple lol ^_^
anonymous
  • anonymous
Yeah, I know ;)
anonymous
  • anonymous
:)
anonymous
  • anonymous
\[|\frac{\ln (n+1))}{n+1}/\frac{\ln(n)}{n}|=\frac{\ln(n+1)}{\ln(n)}\frac{n}{n+1}\]This is an indeterminate form (infty/infty), so you can use L'Hopital's rule:\[\lim_{n->\infty}\frac{\ln (n+1)}{\ln(n)}\frac{n}{n+1}=\lim_{n->\infty}\frac{\ln (n+1)}{\ln (n)}\lim_{n->\infty}\frac{n}{n+1}\]\[=\lim_{n->\infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}\lim_{n->\infty}\frac{1}{1}=\lim_{n->\infty}\frac{n}{n+1}=\lim_{n->\infty}\frac{1}{1+\frac{1}{n}}=1\]

Looking for something else?

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