Question

How do you apply the ratio test to determine if `Sigma (1*3*5* * * (2n-1))/(n!)^2` from `n=[1,oo)` is convergent to divergent?

Answer 1

The series:

`sum_(n=1)^oo (prod_1^n 2k-1)/(n!)^2`

is convergent.

Explanation:

Write the series as:

`sum_(n=1)^oo (prod_1^n 2k-1)/(n!)^2`

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

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

We then evaluate the ratio for the series at hand:

`abs(a_(n+1)/a_n) = frac ((prod_1^(n+1) (2k-1))/((n+1)!^2)) ((prod_1^n (2k-1))/(n!)^2) = ( (prod_1^(n+1) (2k-1)) / (prod_1^n (2k-1)) ) ((n!^2)/((n+1)!^2)) = (2(n+1)-1)/((n+1)^2) = (2n+1)/((n+1)^2)`

so that:

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

which proves the series is convergent.