## megaray26 2 years ago Does this series converge or diverge?

1. megaray26

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

2. amistre64

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

3. megaray26

oops i meant k = 1

4. amistre64

:) 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

5. amistre64

have you tried a ratio test?

6. megaray26

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.

7. amistre64

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

8. amistre64

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

9. megaray26

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.

10. amistre64

a calculator would determine the bigger or less than stuff maybe $\frac{1}{(k+1)^3}-\frac{ k \ln k }{ (k + 1)^3 }>0$ $1-k \ln k>0$ $1>k \ln k$

11. amistre64

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

12. amistre64

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

13. amistre64

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

14. exraven

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