AP Calculus BC

Does the Sum of (n!)^2/(2n)! Converge?

The sum of n factorial squared over 2n factorial converges. The ratio test gives a limit of one quarter, which is less than 1. The key step is that 2n plus 2 factorial equals 2n plus 2 times 2n plus 1 times 2n factorial, so two factors come out rather than one.

∑n=1∞(n!)2(2n)!\sum_{n=1}^{\infty}\frac{(n!)^{2}}{(2n)!}

Converges

Settled by the ratio test.

Expanding the doubled factorial

The step that decides this problem is recognising that going from (2n)!(2n)! to (2n+2)!(2n+2)! picks up TWO factors, not one.

(2n+2)!=(2n+2)(2n+1)(2n)!(2n+2)! = (2n+2)(2n+1)(2n)!
an+1an=((n+1)!)2(2n+2)!⋅(2n)!(n!)2=(n+1)2(2n+2)(2n+1)\frac{a_{n+1}}{a_{n}} = \frac{\left((n+1)!\right)^{2}}{(2n+2)!}\cdot\frac{(2n)!}{(n!)^{2}} = \frac{(n+1)^{2}}{(2n+2)(2n+1)}
L=lim⁡n→∞n2+2n+14n2+6n+2=14<1L = \lim_{n \to \infty}\frac{n^{2}+2n+1}{4n^{2}+6n+2} = \frac{1}{4} < 1

The mistakes students make

  • Writing (2n+2)!=(2n+2)(2n)!(2n+2)! = (2n+2)(2n)!, picking up one factor instead of two. That single slip changes LL from 14\frac{1}{4} to 12\frac{1}{2}, and on a harder problem it would flip the verdict.
  • Reading (2n)!(2n)! as 2⋅n!2 \cdot n!. They are wildly different: (2⋅3)!=720(2 \cdot 3)! = 720 while 2⋅3!=122 \cdot 3! = 12.
  • Reading (n!)2(n!)^{2} as (n2)!(n^{2})!.

Not sure which test a series wants?

The Convergence Test Chooser walks the decision in order: nth term first, then geometric and p-series pattern matching, then alternating structure, then the ratio test, and finally the comparison family.

Frequently asked questions

Does the sum of (n!)^2/(2n)! converge?

Yes. The ratio test gives L=14<1L = \frac{1}{4} < 1.

Why does (2n+2)! give two factors?

Because the factorial steps down one integer at a time: (2n+2)!=(2n+2)(2n+1)(2n)!(2n+2)! = (2n+2)(2n+1)(2n)!. Going from 2n2n to 2n+22n+2 passes two integers.

Is (2n)! the same as 2(n!)?

No. At n=3n = 3, (2n)!=720(2n)! = 720 and 2(n!)=122(n!) = 12.