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manpreet Best Answer 3 years ago

I have this question from a friend who is taking college admission exam, evaluate:

lim n \to \infty ( 4 n 2 n ) 4 n ( 2 n n )

The only way I could do this is by using Stirling's formula:

n! \sim 2 π n - - - \sqrt (n e) n

after rewriting as

lim n \to \infty ( 4 n ) ! ( n ! ) 2 4 n ( 2 n ) ! 3

and it simplifies really satisfying to n id="MathJax-Element-5-Frame" class="MathJax" style="margin: 0px; padding: 0px; b

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manpreet 3 years ago

l = lim n \to \infty ( 4 n ) ! ( n ! ) 2 4 n ( 2 n ) ! 3 = lim n \to \infty n ! \cdot ( 2 \cdot 2 n ) ( 4 n - 1 ) ( 2 \cdot ( 2 n - 1 ) ) . . . ( 2 n + 1 ) 2 n 2 n ( 2 n ) ! \cdot ( 2 n ) ( 2 n - 1 ) . . . ( n + 1 )

And with some adjustments results in:

l = lim n \to \infty n">spa n

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Limit involving binomial coefficients without Stirling's formula

Manpreet Singh

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manpreet Best Answer 3 years ago

manpreet 3 years ago

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