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megaray26 Group TitleBest ResponseYou've already chosen the best response.0
\[\sum_{n = 1}^{\infty} \frac{ k \ln k }{ (k + 1)^3 }\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
there is no n in it, si its just a big ole constant
 one year ago

megaray26 Group TitleBest ResponseYou've already chosen the best response.0
oops i meant k = 1
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
:) 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
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
have you tried a ratio test?
 one year ago

megaray26 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
think we would compare it to 1/k^3 ? or 1/(k+1)^3?
 one year ago

megaray26 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
a calculator would determine the bigger or less than stuff maybe \[\frac{1}{(k+1)^3}\frac{ k \ln k }{ (k + 1)^3 }>0\] \[1k \ln k>0\] \[1>k \ln k\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{k}{(k+1)^3}\frac{ k \ln k }{ (k + 1)^3 }>0\] \[kk \ln k>0\] \[k(1 \ln k)>0\] might be able to work out something with that
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
when k=1, 1(10)>0 when k=e, e(11)=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
 one year ago

exraven Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago
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