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Question Number 88761 by M±th+et£s last updated on 12/Apr/20

i f a_{n} = \frac{n!}{n^{n} e^{- n} \sqrt{2 π n}}

a n d b_{n} = \frac{(2 n)! \sqrt{n}}{4^{n} {(n!)}^{2}}

Double subscripts: use braces to clarify

Double subscripts: use braces to clarify

Commented by M±th+et£s last updated on 12/Apr/20

i n e e d a h e l p p l e a s e . .

Commented by abdomathmax last updated on 12/Apr/20

s t r l i n g f o r m i l a e n! \sim n^{n} e^{- n} \sqrt{2 π n} (n \to + \infty) \Rightarrow

(2 n)! \sim {(2 n)}^{2 n} e^{- 2 n} \sqrt{4 π n} \Rightarrow

b_{n} \sim \frac{{(2 n)}^{2 n} e^{- 2 n} \sqrt{4 π n} \times \sqrt{n}}{4^{n} n^{2 n} e^{- 2 n} (2 π n)}

= \frac{2 \sqrt{π} n}{(2 π n)} = \frac{1}{\sqrt{π}} \Rightarrow {l i m}_{n \to + \infty} b_{n} = \frac{1}{\sqrt{π}}

Commented by TANMAY PANACEA. last updated on 12/Apr/20

Commented by M±th+et£s last updated on 12/Apr/20

t h a n x s i r g o d b l e s s y o u