I am working on the limit
limn→∞Γ(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)/(√2πnn−12e−n)
In particular, how do I relate Γ(n) to nn and e−n? Not sure how do deal with those two terms.
Answer
You know that
limn→∞Γ(n)√2πnn−1/2e−n=1.
It seems to me that this is how Γ(n) relates to
nn and e−n.
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
limn→∞Γ(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
limn→∞g(n)f(n)?
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