Root the middle Binomial

Calculus Level 3

lim ⁡ n → ∞ ( 2 n n ) n \lim_{n \rightarrow \infty} \sqrt[n] { 2n \choose n }

e e 4 4 0 0 π \pi

Calvin Lin by Calvin Lin

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2 solutions

Michael Mendrin
Oct 3, 2015

Using the good old Sterling's Approximation gets you the answer

( 2 π 2 n ( 2 n e ) 2 n 2 π n ( n e ) n 2 π n ( n e ) n ) 1 n = 4 ⋅ 1 ( π n ) 1 n { \left( \dfrac { \sqrt { 2\pi 2n } { \left( \frac { 2n }{ e } \right) }^{ 2n } }{ \sqrt { 2\pi n } { \left( \frac { n }{ e } \right) }^{ n }\sqrt { 2\pi n } { \left( \frac { n }{ e } \right) }^{ n } } \right) }^{ \frac { 1 }{ n } }=4\cdot \frac { 1 }{ { \left( \sqrt { \pi n } \right) }^{ \frac { 1 }{ n } } }

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Of course, that's not surprising.

How can we approach this otherwise?

Calvin Lin Staff - 5 years, 8 months ago

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I kind of give up. I keep coming back to the Wallis infinite product as an alternate way of coming up with this result. Here’s how it’s done if we go that way

( 2 n ) ! n ! n ! = \dfrac { \left( 2n \right) ! }{ n!n! } =

2 n ( 2 n − 1 ) ! ! n ! = \dfrac { { 2 }^{ n }\left( 2n-1 \right) !! }{ n! } =

4 n 1 2 n + 1 ∏ k = 1 n ( 2 k − 1 2 k ⋅ 2 k + 1 2 k ) { 4 }^{ n }\sqrt { \dfrac { 1 }{ 2n+1 }\displaystyle \prod _{ k=1 }^{ n }{ \left( \dfrac { 2k-1 }{ 2k } \cdot \dfrac { 2k+1 }{ 2k } \right) } }

which of course is a variation of the Wallis infinite product, with the limit value of

4 n 1 n π { 4 }^{ n }\dfrac { 1 }{ \sqrt { n\pi } }

and the rest follows from this. I’m not sure why I can’t seem to find a more obvious way to prove this, without invoking either Stirling’s approximation or the Wallis infinite product.

Michael Mendrin - 5 years, 8 months ago

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...oh, wait, now I feel kind of stupid. Let's expand ( 1 + 1 ) 2 n { \left( 1+1 \right) }^{ 2n } to display all the coefficients. The center coefficient, out of 2 n + 1 2n+1 terms, is the largest. So we can infer that

2 2 n > ( 2 n ) ! n ! n ! > 2 2 n 2 n + 1 { 2 }^{ 2n } > \dfrac { \left( 2n \right) ! }{ n!n! } > \dfrac { { 2 }^{ 2n } }{ 2n+1 }

I think we can deduce the limit value from this. Let me think on it. ...Yeah, just take the n t h nth root of all terms, and we're done.

Michael Mendrin - 5 years, 8 months ago

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@Michael Mendrin – Yup, that's the approach using less machinery.

Calvin Lin Staff - 5 years, 8 months ago

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@Calvin Lin – It's the short, "elegant" proofs that can take the longest time to come up with. This was somehow a stranger problem than I thought it'd be.

Michael Mendrin - 5 years, 8 months ago

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@Michael Mendrin – l'm just here to see how someone got this. I used an android app called Symbolab. LOL.

E Tyson Ewing III - 5 years, 7 months ago
James Wilson
Nov 15, 2017

After taking the natural logarithm, I was able to put it into the form of a Riemann integral, and take the antiderivative to get the answer. Let L L be the desired limit. Then ln ⁡ L = lim ⁡ n → ∞ 1 n ln ⁡ ( 2 n ) ! n ! n ! = lim ⁡ n → ∞ 1 n ln ⁡ 2 n ( 2 n − 1 ) . . . ( n + 1 ) n ( n − 1 ) . . . ( 1 ) \ln{L}=\lim_{n\rightarrow \infty} \frac{1}{n}\ln{\frac{(2n)!}{n!n!}}=\lim_{n\rightarrow \infty} \frac{1}{n} \ln{\frac{2n(2n-1)...(n+1)}{n(n-1)...(1)}} = lim ⁡ n → ∞ 1 n ∑ k = 0 ∞ ln ⁡ 2 n − k n − k = ∑ k = 0 ∞ ln ⁡ 2 − k / n 1 − k / n = ∫ 0 1 ln ⁡ 2 − x 1 − x d x = ∫ 0 1 ln ⁡ x + 1 x d x =\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=0}^\infty \ln{\frac{2n-k}{n-k}}=\sum_{k=0}^\infty \ln{\frac{2-k/n}{1-k/n}}=\int_0^1 \ln{\frac{2-x}{1-x}}dx= \int_0^1 \ln{\frac{x+1}{x}}dx = ∫ 0 1 ( ln ⁡ ( x + 1 ) − ln ⁡ x ) d x = [ ( x + 1 ) ln ⁡ ( x + 1 ) − x ln ⁡ x ] 0 1 = 2 ln ⁡ 2 − 0 − 0 + lim ⁡ x → 0 + x ln ⁡ x = 2 ln ⁡ 2 =\int_0^1 (\ln{(x+1)}-\ln{x})dx =\Big[(x+1)\ln{(x+1)}-x\ln{x}\Big]_0^1=2\ln{2}-0-0+\lim_{x\rightarrow 0+}x\ln{x}=2\ln{2} . Therefore, L = exp ⁡ ( 2 ln ⁡ 2 ) = 4 L=\exp(2\ln{2})=4 .

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