Tuesday, December 29, 2015

calculus - Evaluate limnrightarrowinftyGamma(n+frac12)/left(sqrt2npiGamma(n)right) using Stirling's formula.



I am working on the limit
limnΓ(n+12)2nπΓ(n).



I am thinking I may be able to use Stirling's formula, but they are slightly different, and I am having trouble relating the two. Any help is appreciated.




Stirling's formula says that the limit is 1 as n approaches infinity of the following:



Γ(n)/(2πnn12en)



In particular, how do I relate Γ(n) to nn and en? Not sure how do deal with those two terms.


Answer



You know that
limnΓ(n)2πnn1/2en=1.



It seems to me that this is how Γ(n) relates to
nn and en.
What you need to know, though, is whether this will help you
relate Γ(n+12) to Γ(n)2nπ,
and if it does, how does it?



Notice what happens if we replace n by n+12 in Equation (1).
We get

limnΓ(n+12)2π(n+12)ne(n+1/2)=1.



Let f(n) be the left-hand side of (1)
and g(n) be the left-hand side of (2).
What can you say about
limng(n)f(n)?


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