# For which values of a does this series converge?

1. May 17, 2010

### ziggie125

[/PHP]1. The problem statement, all variables and given/known data

For which values of a does this series converge?

$$\sum$$ (n!)^2/(an)!

3. The attempt at a solution

I know a cannot be a negative integer because you cannot have a negative factorial.

If a is 0, then it's limit is infinity. ie. lim (n!)^2/ 0 = infinity

If a is +1, then lim cn+1/cn = ((n+1)!)^2/(n+1)! x n!/(n!)^2

= lim (n+1)^2/(n+1) = lim (n+1)/1 = infinity

For the series to converge it's limit from n to infinity must be equal to 0 right? So is there any value of a where it converges, or is my math wrong?

2. May 17, 2010

### physicsman2

If a = 0, 0! = 1

3. May 17, 2010

### ziggie125

thx. The limit is still infinity though.

4. May 17, 2010

### phyzguy

Try using Stirling's formula. I think you'll see then that the series does converge as long as a is greater than some number.