Question

How do you apply the ratio test to determine if `Sigma (4^n(n!)^2)/((2n)!)` from `n=[1,oo)` is convergent to divergent?

Answer 1

The ratio test is indeterminate for this series.

Explanation:

For the series:

`sum_(n=1)^oo a_n = sum_(n=1)^oo (4^n(n!)^2)/((2n)!)`

Evaluate the ratio:

`abs(a_(n+1)/a_n) = abs ( ( (4^(n+1)((n+1)!)^2)/((2(n+1))!) ) / ((4^n(n!)^2)/((2n)!)))`

`abs(a_(n+1)/a_n) = 4^(n+1)/4^n ((n+1)!)^2/ ((n!)^2) ((2n)!)/ ((2n+2)!)`

`abs(a_(n+1)/a_n) = (4 (n+1)^2) / ((2n+2)(2n+1))`

`abs(a_(n+1)/a_n) = (4n^2+8n+4) / (4n^2+6n+2)`

So we have:

`lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo)(4n^2+8n+4) / (4n^2+6n+2) = 1`

which means the ratio test is indecisive and we cannot determine whether the series is convergent or not.