I'm looking for a proof of the following limit:
limn→∞enn!nn√n=√2π
This follows from Stirling's Formula, but how can it be proven?
Answer
Existence of the Limit
The ratio of two consecutive terms is
enn!nn+1/2/en−1(n−1)!(n−1)n−1/2=e(1−1n)n−1/2
Using the Taylor approximation
log(1−1n)=−1n−12n2−13n3+O(1n4)
we can compute the logarithm of the right hand side of (1):
1+(n−12)log(1−1n)=−112n2+O(1n3)
Therefore,
limn→∞enn!nn+1/2=e∞∏n=2(enn!nn+1/2/en−1(n−1)!(n−1)n−1/2)=e∞∏n=2(e(1−1n)n−1/2)=exp(1+∞∑n=2(1+(n−12)log(1−1n)))=exp(1+∞∑n=2(−112n2+O(1n3)))
Since the sum on the right hand side of (4) converges, so does the limit on the left hand side.
Log-Convexity of Γ(x)
Since Γ(x) is log-convex,
√nΓ(n+1/2)≤√nΓ(n)1/2Γ(n+1)1/2=Γ(n+1)
and
Γ(n+1)≤Γ(n+1/2)1/2Γ(n+3/2)1/2=√n+1/2Γ(n+1/2)
Therefore,
√nn+1/2≤√nΓ(n+1/2)Γ(n+1)≤1
Thus, by the Squeeze Theorem,
limn→∞√nΓ(n+1/2)Γ(n+1)=1
Using the Recursion for Γ(x)
Using the identity Γ(x+1)=xΓ(x),
√n(2n)!2nn!12nn!=√n12⋅34⋅56⋯2n−12n=√n1/21⋅3/22⋅5/23⋯n−1/2n=√nΓ(n+1/2)Γ(1/2)Γ(1)Γ(n+1)
Value of the Limit
Combining (8) and (9) yields
limn→∞(enn!nn+1/2)−1=limn→∞e2n(2n)!(2n)2n+1/2(nn+1/2enn!)2=limn→∞√n√24n(2n)!(n!)2=limn→∞1√2Γ(1)Γ(1/2)√nΓ(n+1/2)Γ(n+1)=1√2π
Therefore,
limn→∞enn!nn+1/2=√2π
No comments:
Post a Comment