3.2 Continuity, Derivatives, and Integrals 163To bring this closer to Stirling’s formula, note that the term in the middle is equal to
e−n−^1 (n+ 1 )n+^1 ((n+ 1 )!)−^1√
n+ 1
e−nnn(n!)−^1√
n=
xn+ 1
xn,
wherexn=e−nnnn!
√
n, a number that we want to prove is equal to√
2 πe−θn
12 n with
0 <θn<1. In order to prove this, we write the above double inequality as
1 ≤
xn
xn+ 1≤
e
121 n
e
12 (n^1 + 1 ).We deduce that the sequencexnis positive and decreasing, while the sequencee−
121 n
xnis
increasing. Becausee−
121 n
converges to 1, and because(xn)nconverges by the Weierstrass
criterion, bothxnande−
121 n
xnmust converge to the same limitL. We claim thatL=
√
2 π.
Before proving this, note that
e−
121 n
xn<L<xn,so by the intermediate value property there existsθn∈( 0 , 1 )such thatL=e−
12 θnn
xn, i.e.,
xn=e
12 θnn
L.
The only thing left is the computation of the limitL. For this we employ the Wallis
formula
nlim→∞[
2 · 4 · 6 ··· 2 n
1 · 3 · 5 ···( 2 n− 1 )] 2
1
n=π,proved in problem 466 from Section 3.2.8 (the one on definite integrals). We rewrite this
limit as
lim
n→∞22 n(n!)^2
( 2 n)!·
1
√
n=
√
π.Substitutingn!and( 2 n)!by the formula found above gives
nlim→∞nL^2(n
e) 2 n
e
212 θnn
22 n
√
2 nL( 2 n
e) 2 n
eθ 2 n
24 n·
1
√
n=nlim→∞1
√
2
Le4 θn−θ 2 n
24 n =√
π.HenceL=
√
2 π, and Stirling’s formula is proved.
Try your hand at the following problems.486.Prove that for any real numberx, the series
1 +x^4
4!+
x^8
8!+
x^12
12!+···
is convergent and find its limit.