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anonymous
 5 years ago
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
anonymous
 5 years ago
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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...why the ratio test? painful.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The ratio I get using the ratio test is 1. It wouldn't be a conclusive test.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you want me to show you that, let me know.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's not that painful loki all you've got to do is put : an+1 / an , I find it rather simple lol ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\ln (n+1))}{n+1}/\frac{\ln(n)}{n}=\frac{\ln(n+1)}{\ln(n)}\frac{n}{n+1}\]This is an indeterminate form (infty/infty), so you can use L'Hopital's rule:\[\lim_{n>\infty}\frac{\ln (n+1)}{\ln(n)}\frac{n}{n+1}=\lim_{n>\infty}\frac{\ln (n+1)}{\ln (n)}\lim_{n>\infty}\frac{n}{n+1}\]\[=\lim_{n>\infty}\frac{\frac{1}{n+1}}{\frac{1}{n}}\lim_{n>\infty}\frac{1}{1}=\lim_{n>\infty}\frac{n}{n+1}=\lim_{n>\infty}\frac{1}{1+\frac{1}{n}}=1\]
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