AP Calculus BC

Does the Sum of (2n)!/(n!)^2 Converge? No

The series diverges. The ratio test gives a limit of 4, which is greater than one, so the terms eventually multiply by nearly four at every step. These terms are the central binomial coefficients, and they grow rather than shrink.

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

Diverges

Settled by the ratio test.

Cancelling the factorials

an+1an=(2n+2)!((n+1)!)2⋅(n!)2(2n)!=(2n+1)(2n+2)(n+1)2\frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{((n+1)!)^{2}} \cdot \frac{(n!)^{2}}{(2n)!} = \frac{(2n+1)(2n+2)}{(n+1)^{2}}

Expanding the top gives 2(2n+1)(n+1)2(2n+1)(n+1), so the quotient is 2(2n+1)n+1\frac{2(2n+1)}{n+1}, which tends to 4. Since 4>14 > 1, the ratio test proves divergence.

Never expand a factorial. Write (2n+2)!=(2n+2)(2n+1)(2n)!(2n+2)! = (2n+2)(2n+1)(2n)! and cancel; the whole method depends on peeling off exactly the factors that differ.

What these terms are

The term (2n)!(n!)2\frac{(2n)!}{(n!)^{2}} is the binomial coefficient (2nn)\binom{2n}{n}, the middle entry of row 2n2n of Pascal's triangle. It counts the ways to choose nn items from 2n2n, so it is a whole number and it is large: (2010)=184756\binom{20}{10} = 184756.

Growing terms mean the nth term test would also have settled this, and faster. The ratio test is still worth running because it names the growth factor, and the same computation is what finds the radius of convergence when a power series is built on these coefficients.

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

Could the nth term test have decided this?

Yes, and in one line: the terms grow without bound, so they do not tend to 0. Always try the nth term test first; it costs one limit and often finishes the problem.

How do I simplify a ratio of factorials?

Write the larger factorial in terms of the smaller one and cancel. (n+1)!=(n+1) n!(n+1)! = (n+1)\,n! and (2n+2)!=(2n+2)(2n+1)(2n)!(2n+2)! = (2n+2)(2n+1)(2n)! are the two you need most.