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
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

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.