Wednesday, 25 December 2013

limits - Proof of limntoinftyfracenn!nnsqrtn=sqrt2pi



I'm looking for a proof of the following limit:

limnenn!nnn=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/en1(n1)!(n1)n1/2=e(11n)n1/2



Using the Taylor approximation
log(11n)=1n12n213n3+O(1n4)

we can compute the logarithm of the right hand side of (1):
1+(n12)log(11n)=112n2+O(1n3)

Therefore,

limnenn!nn+1/2=en=2(enn!nn+1/2/en1(n1)!(n1)n1/2)=en=2(e(11n)n1/2)=exp(1+n=2(1+(n12)log(11n)))=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+1/2nΓ(n+1/2)Γ(n+1)1

Thus, by the Squeeze Theorem,
limnnΓ(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!=n1234562n12n=n1/213/225/23n1/2n=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=limne2n(2n)!(2n)2n+1/2(nn+1/2enn!)2=limnn24n(2n)!(n!)2=limn12Γ(1)Γ(1/2)nΓ(n+1/2)Γ(n+1)=12π


Therefore,
limnenn!nn+1/2=2π


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