Question

Does this series converge or diverge?

Answer 1

\[\sum_{n = 1}^{\infty} \frac{ k \ln k }{ (k + 1)^3 }\]

Answer 2

there is no n in it, si its just a big ole constant

Answer 3

oops i meant k = 1

Answer 4

:) my gut says taht the bottom is always bigger than the top and I would want to say it converges eventually. but I got no tests in mind to prove that

Answer 5

have you tried a ratio test?

Answer 6

Well i have, but it doesn't look like it will lead to an easier solution. It just seemed to make the problem harder, unless there is some trick that I am missing while doing so.

Answer 7

http://www.wolframalpha.com/input/?i=sum+k*ln%28k%29%2F%28k%2B1%29%5E3%2C+k%3D1..inf

says a comparison test has it converging

Answer 8

think we would compare it to 1/k^3 ? or 1/(k+1)^3?

Answer 9

What i had thought of using was comparing to k / k^3, or 1/ k^2, and maybe try a limit comparison test, but that didn't exactly work out.
Also, for those, I'm not exactly sure if the series is less than 1 / k^3, which is what I need for convergence, but I may be wrong.

Answer 10

a calculator would determine the bigger or less than stuff maybe

\[\frac{1}{(k+1)^3}-\frac{ k \ln k }{ (k + 1)^3 }>0\]
\[1-k \ln k>0\]
\[1>k \ln k\]

Answer 11

if we do a k up top we might be able to factor out a k first, but most of this is more art than science to me

Answer 12

\[\frac{k}{(k+1)^3}-\frac{ k \ln k }{ (k + 1)^3 }>0\]
\[k-k \ln k>0\]
\[k(1- \ln k)>0\] might be able to work out something with that

Answer 13

when k=1, 1(1-0)>0
when k=e, e(1-1)=0

http://www.wolframalpha.com/input/?i=y%3D1%2F%28x%2B1%29%5E3%2C+y%3Dx*ln%28x%29%2F%28x%2B1%29%5E3

hmm, the comparison seems to be smaller in both cases than the function we are looking at

Answer 14

the limit comparison test works, try\[1/k^{\frac{3}{2}}\]you will get\[\lim_{k \rightarrow \infty} \frac{k^{\frac{5}{2}}\ln k}{(k + 1)^3} = 0\]therefore the series converges