- #1
ziggie125
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For which values of a does this series converge?
[tex]\sum[/tex] (n!)^2/(an)!
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?
Homework Statement
For which values of a does this series converge?
[tex]\sum[/tex] (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?