How can I compute the following limit?
limn→∞1nn(−Γ(n)n+enΓ(n+1,n))√n
The answer appears to be about 1.25.
Answer
This is the same as the one worked out in this question. We have
Γ(n+1,n)=n!en(n∑k=0nkk!)∼n!2
from this question.
Hence, we get that
enΓ(n+1,n)nn+1/2∼enn!2nn+1/2∼en√2πnn+1/22nn+1/2en=√π2
As I have in the comments, the first term can be thrown away. Hence, the limit is √π2≈1.25
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