SolomonZelman
  • SolomonZelman
I'm looking at the following series (for fun), and I this is what I got....
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!
SolomonZelman
  • SolomonZelman
\(\large\color{#000000}{ \displaystyle \sum_{ n=2 }^{ \infty } ~ \frac{ 1}{\ln\Gamma(kx)}}\) When \(k<0\) and \(k\notin \mathbb{Z}\) I did the limit test and the limit (and thus the series) diverges (does not approach a value). (I know if it converged, no colnclusions...) When \(k\le0\) and \(k\in \mathbb{Z}\) Then all terms are zeros and series converges.
SolomonZelman
  • SolomonZelman
\(\color{#000000}{ \displaystyle \sum_{ n=2 }^{ \infty } ~ \frac{ 1}{\ln\Gamma(kx)}}\) When \(k\in \mathbb{N}\). We can show divergence by the comparison Test. \(\color{#000000 }{ \displaystyle \frac{1}{\ln\Gamma (kx)}>\frac{1}{\ln (kx{\tiny~}!)}>\frac{1}{\ln \left[(kx)^{kx}\right] } }\) \(\color{#000000 }{ \displaystyle \frac{1}{\ln\Gamma (kx)}>\frac{1}{kx\ln \left[kx\right] } }\) Therefore, \(\color{#000000}{ \displaystyle \sum_{ n=2 }^{ \infty } ~ \frac{ 1}{\ln\Gamma(kx)}>\sum_{ n=2 }^{ \infty } ~ \frac{1}{kx\ln \left[kx\right] }}\) And since the series on the right diverges. (When integrated, I got (1/k)•ln\((\)ln(kx)\()\), and the limit of this as \(x\to\infty\) will diverge to ∞.)
SolomonZelman
  • SolomonZelman
When \(k\notin \mathbb{N}\) and \(k>0\). We can show that the series converges by the squeeze theorem. For example, we know that \(\color{#000000}{ \displaystyle \sum_{ n=2 }^{ \infty } ~ \frac{ 1}{\ln\Gamma(\color{red}{100}x)}}\) and \(\color{#000000}{ \displaystyle \sum_{ n=2 }^{ \infty } ~ \frac{ 1}{\ln\Gamma(\color{red}{101}x)}}\) both diverge. Therefore, (for instance), \(\color{#000000}{ \displaystyle \sum_{ n=2 }^{ \infty } ~ \frac{ 1}{\ln\Gamma(\color{red}{100}x)}>\sum_{ n=2 }^{ \infty } ~ \frac{ 1}{\ln\Gamma(\color{blue}{100.6}x)}>\sum_{ n=2 }^{ \infty } ~ \frac{ 1}{\ln\Gamma(\color{red}{101}x)}}\) And since the left and right diverge (because they are just the same case of \(k\in\mathbb{N}\)), therefore the middle series diverges by the Squeeze Theorem.

Looking for something else?

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

More answers

SolomonZelman
  • SolomonZelman
For \(0\frac{1}{\ln \left[(kx)^{kx}\right] } }\) (This would be true if not at n=2, then for sure from some n=j and on, and if we show that series diverges from n=j then it would diverge from n=2 too) So, in this case; \(0
benlindquist
  • benlindquist
wt
anonymous
  • anonymous
umm....
anonymous
  • anonymous
hes either a rocket scientist or a whale bioligist(i know what im doing im a whale bioligist)
anonymous
  • anonymous
Maybe a silly question, but since your index is n, I'm assuming you intentionally do not have an n in the summand, or is it missing? If we are not supposed to have n in the summand, then we're really only considering \[\lim_{n \rightarrow \infty} \frac{ n }{ \ln \Gamma (kx) }\] Am I missing something?
SolomonZelman
  • SolomonZelman
Yes, I apologize, that is supposed to be nx.
SolomonZelman
  • SolomonZelman
Or else the unknown constant is the product of k and x, rather than just k. Thanks for asking.
anonymous
  • anonymous
yep whale bioligist

Looking for something else?

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