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

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

amistre64
 one year ago
Best 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

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

megaray26
 one year ago
Best ResponseYou've already chosen the best response.0Well 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.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1http://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

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

megaray26
 one year ago
Best ResponseYou've already chosen the best response.0What 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.

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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1if 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

amistre64
 one year ago
Best 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1when 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

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