Find all the values of the p>0 for which the series [sigma(n!)^p/(3n)! from 1 to infinity] is convergent?

Question

Find all the values of the constant p>0 for which the fallowing series is convergent:
the sum of the series (n!)^p/(3n)! from 1 to infinity

Answer 1

Consider the ratio of the nth term to the (n-1)th term:

n^p/((3n-2)(3n-1)(3n))

As n approaches infinity, the lower order terms drop out, and so we have:

[n→∞]lim |n^p/((3n-2)(3n-1)(3n))| = [n→∞]lim |n^p/(27n³)|

If p<3, then this limit is 0, so the series converges, by the ratio test.
If p=3, then this limit is 1/27, which is less than 1, so the series also converges by the ratio test
If p>3, then this limit is ∞, so the series diverges by the ratio test.

Therefore, the series converges for all p≤3.

Answer 2

at a guess it must be p<3

a[n] = (n!)^p / (3n!)
a[n+1]/a[n] =(n+1)^p / [(3n+3)(3n+2)(3n+1)
= (1/27) (n^p/n^3) (1+1/n)^p / [(1+1/n)(1+2/(3n))(1+1/(3n))
< (1/27)(1+1/n)^p -->(1/27) if p<=3

if p>3 ratio
> (1/27) (n^p/n^3) (1+1/n)^p /[(1+1/n)(1+1/n)(1+1/n)
which diverges, so the series must diverge

so convergence requires p<= 3

(my guess slightly wrong!)

Answer 3

I too think of this is convergent. because of the fact the fee of the words will shrink at one factor and the sum could be a finite quantity. yet, I remember from someplace that if r is any term interior the series and r is often reducing then the series is convergent. If r is increasing then it could be divergent.