Values of k s.t. the series converges

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SUMMARY

The series \(\sum_{n=1}^{\infty}{\frac{(n!)^2}{(kn)!}}\) converges for positive integers \(k\) greater than 2. The analysis involved simplifying the series to \(\sum_{n=1}^{\infty}{\frac{(n!)}{(n+1)(n+2)...(kn-1)(kn)}}\) and applying the limit comparison test. The conclusion drawn is that for convergence, the number of terms in the denominator must exceed those in the numerator, leading to the inequality \(k > 2\). This indicates that while the terms approach zero, further testing is required to confirm convergence.

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Homework Statement


For which positive integers k is the following series convergent?
[tex]\sum_{n=1}^{\infty}{\frac{(n!)^2}{(kn)!}}[/tex]

the latex code is acting weird, i'll put it in regular text too:
Sum (n=1 --> infinity) (n!)2/(kn)!


Homework Equations


To test for convergence, I can use the alternating series test, limit comparison test, comparison test, ratio test, root test


The Attempt at a Solution


I canceled one of the n! on top with part of the (kn)! on the bottom to get:
[tex]\sum_{n=1}^{\infty}{\frac{(n!)}{(n+1)(n+2)...(kn-1)(kn)}}[/tex]

w/o latex:
Sum (n=1 --> infinity) (n)!/[(n+1)(n+2)...(kn-1)(kn)]
I asked another teacher about this (my calc teacher says we can use any resource except him for this problem) and he said that in order for it to converge, there must be more terms on the bottom of the fraction than on top (so the limit as n-->infinity of each term is zero). Therefore, I have n terms on top and kn-n on the bottom, so:
[tex]n\le kn-n[/tex] n=<kn-n
and thus
[tex]2n\le kn[/tex] 2n=<kn
[tex]2\le k[/tex] 2=<k

So k must be greater than 2, but my problem is that I don't think this proves the series converges, merely that the terms approach zero as n gets very large. Is my thinking correct (in which case help please), or does this indeed prove convergence?
 
Last edited:
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Go back to the original unsimplified series, use k=2, and try ratio test.
 

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