The ratio test states that a necessary condition for a series sum_(n=1)^oo a_n∞∑n=1an 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:
- 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.
- 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.
- 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.
