For which values of a does this series converge?

[ziggie125]

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Homework Statement

For which values of a does this series converge?

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

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?

 

[physicsman2]

[/PHP]

Homework Statement

For which values of a does this series converge?

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

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?

If a = 0, 0! = 1

 

[ziggie125]

thx. The limit is still infinity though.

 

[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.