anonymous
  • anonymous
Does this series converge or diverge?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\sum_{n = 1}^{\infty} \frac{ k \ln k }{ (k + 1)^3 }\]
amistre64
  • amistre64
there is no n in it, si its just a big ole constant
anonymous
  • anonymous
oops i meant k = 1

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • 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
amistre64
  • amistre64
have you tried a ratio test?
anonymous
  • anonymous
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.
amistre64
  • 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
amistre64
  • amistre64
think we would compare it to 1/k^3 ? or 1/(k+1)^3?
anonymous
  • anonymous
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.
amistre64
  • amistre64
a calculator would determine the bigger or less than stuff maybe \[\frac{1}{(k+1)^3}-\frac{ k \ln k }{ (k + 1)^3 }>0\] \[1-k \ln k>0\] \[1>k \ln k\]
amistre64
  • 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
amistre64
  • 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
amistre64
  • 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
anonymous
  • anonymous
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.