Question

How do you use the ratio test to test the convergence of the series `∑ (n!)^2 / (kn)!` from n=1 to infinity?

Answer 1

The series:

`sum_(n=1)^oo ((n!)^2)/((kn)!)`

is convergent for `k>=2`

Explanation:

The ratio test states that a necessary condition for a series `sum_(n=1)^oo a_n` to converge is that:

`L= lim_(n->oo) abs (a_(n+1)/a_n) <= 1`

if `L < 1` the condition is also sufficient.

In our case:

`abs (a_(n+1)/a_n) = ((((n+1)!)^2)/((k(n+1))!))/(((n!)^2)/((kn)!)) = (((n+1)!)^2)/((n!)^2) ((kn)!)/ ((kn+k)!) = (n+1)^2/((kn+k) (kn+k-1)...(kn+1))`

Now we have that:

1. For `k= 1`

`lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) (n+1)^2/(n+1) = lim_(n->oo) (n+1) = +oo`

and the series is divergent.

1. For `k= 2`

`lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) (n+1)^2/((n+2)(n+1)) = 1`

and the test is inconclusive, so we have to look at the series in more detail:

`sum_(n=1)^oo ((n!)^2)/((2n)!) = (n(n-1)(n-2)...2*1)/ (2n(2n-1)...(n+1)`

We can note that the numerator has `n` factors from `1` to `n` and the denominator has `n` factors from `n+1` to `2n`, so ordering them appropriately we have:

`(n(n-1)(n-2)...2)/ (2n(2n-1)...(n+1)) = prod_(q=1)^n q/(n+q)`

Now as `q <= n`,

`q/(n+q) <= q/(q+q) = 1/2`

So we have:

`(n(n-1)(n-2)...2)/ (2n(2n-1)...(n+1)) <= (1/2)^n`

And as:

`sum_(n=1)^oo (1/2)^n = 1`

is convergent, then also our series is convergent by direct comparison.

1. For `k >= 3`

`lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) (n+1)^2/((n+1)(n+2)...(n+k)) = 0`

and the series is convergent.