Values of k s.t. the series converges

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Homework Statement


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

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:
\sum_{n=1}^{\infty}{\frac{(n!)}{(n+1)(n+2)...(kn-1)(kn)}}

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:
n\le kn-n n=<kn-n
and thus
2n\le kn 2n=<kn
2\le k 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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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