Question

I'm looking at the following series (for fun), and I this is what I got....

Answer 1

\(\large\color{#000000}{ \displaystyle \sum_{ n=2 }^{ \infty } ~ \frac{ 1}{\ln\Gamma(kx)}}\)

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.

Answer 2

\(\color{#000000}{ \displaystyle \sum_{ n=2 }^{ \infty } ~ \frac{ 1}{\ln\Gamma(kx)}}\)

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 ∞.)

Answer 3

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.

Answer 4

For \(0<k<1\), and if \(k\notin\mathbb{Z}\).
We can use the Comprison Test as well.

\(\color{#000000 }{ \displaystyle \frac{1}{\ln\Gamma (kx)}>\frac{1}{\ln \left[(kx)^{kx}\right] } }\)
(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; \(0<k<1\), and if \(k\notin\mathbb{Z}\), series diverges.

Answer 5

wt

Answer 6

umm....

Answer 7

hes either a rocket scientist or a whale bioligist(i know what im doing im a whale bioligist)

Answer 8

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?

Answer 9

Yes, I apologize, that is supposed to be nx.

Answer 10

Or else the unknown constant is the product of k and x, rather than just k. Thanks for asking.

Answer 11

yep whale bioligist