Real Analysis Proof?

Question

A theorem states the following:

(a) If {s_n} is nondecreasing, then lim[n-->infinity]s_n=sup{s_n}.

(b) If {s_n} is nonincreasing, then lim[n-->infinity]s_n=inf{s_n}.

I am not sure how to prove that {s_n} converges if s_n = (2n) ! / {[2^(2n)]*(n !)^2}. Can someone please show me a proof for this problem?

Answer 1

Note first that {s_n} has nonnegative terms.

Look at the ratio of successive terms, s_{n + 1} / s_n:

s_{n + 1} / s_n

= [(2(n + 1))! / {[2^(2(n + 1))] * [(n + 1)!]^2}] / [(2n)! / {[2^(2n)] * (n!)^2}]

= [(2n + 2)! / {[2^(2n + 2)] * [(n + 1)!]^2}] / [(2n)! / {[2^(2n)] * (n!)^2}]

= (2n + 2)(2n + 1) / [2^2 * (n + 1)^2]

= (4n^2 + 6n + 2) / (4n^2 + 8n + 4).

If n > 0, then surely 4n^2 + 8n + 4 > 4n^2 + 6n + 2. So s_{n + 1} / s_n < 1 for all n.

So s_{n + 1} < s_n for all n; that is, {s_n} is nonincreasing.

{s_n} is also obviously bounded below by 0.

Since {s_n} is nonincreasing and bounded below, then it is convergent.