Interesting Limit (6)

Calculus Level 2

lim n [ ( n + 1 ) ! n + 1 n ! n ] = ? \large\lim_{n\to\infty}\left[\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right]=\, ?

2 e \frac 2e e e 1 e \frac 1e 1

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1 solution

Chew-Seong Cheong
Mar 24, 2018

Relevant wiki: Stirling's Formula

L = lim n [ ( n + 1 ) ! n + 1 n ! n ] By Stirling’s formula: n ! 2 π n ( n e ) n = lim n [ 2 π ( n + 1 ) ( n + 1 e ) n + 1 n + 1 2 π n ( n e ) n n ] = lim n [ ( 2 π ( n + 1 ) ) 1 2 n + 2 ( n + 1 e ) ( 2 π n ) 1 2 n ( n e ) ] = lim n 1 e ( ( 2 π ) 1 2 n + 2 ( n + 1 ) 2 n + 3 2 n + 2 ( 2 π ) 1 2 n n 2 n + 1 2 n ) = lim n 1 e ( ( 2 π ) 1 2 n + 2 ( n + 1 ) 1 + 3 2 n 1 + 1 n ( 2 π ) 1 2 n n 1 + 1 2 n ) = 1 e ( n + 1 n ) = 1 e \begin{aligned} L & = \lim_{n \to \infty} \left[\sqrt[n+1]{(n+1)!} - \sqrt[n]{n!} \right] & \small \color{#3D99F6} \text{By Stirling's formula: }n! \thicksim \sqrt{2\pi n}\left(\frac ne \right)^n \\ & = \lim_{n \to \infty} \left[\sqrt[n+1]{\sqrt{2\pi (n+1)}\left(\frac {n+1}e \right)^{n+1}} - \sqrt[n]{\sqrt{2\pi n}\left(\frac ne \right)^n} \right] \\ & = \lim_{n \to \infty} \left[\left(2\pi (n+1)\right)^{\frac 1{2n+2}}\left(\frac {n+1}e \right) - \left(2\pi n\right)^{\frac 1{2n}}\left(\frac ne \right) \right] \\ & = \lim_{n \to \infty} \frac 1e \left((2\pi)^{\frac 1{2n+2}}(n+1)^{\frac {2n+3}{2n+2}} - (2\pi)^{\frac 1{2n}} n^{\frac {2n+1}{2n}} \right) \\ & = \lim_{n \to \infty} \frac 1e \bigg((2\pi)^{\frac 1{2n+2}}(n+1)^{\frac {1+\frac 3{2n}}{1+\frac 1n}} - (2\pi)^{\frac 1{2n}} n^{1+\frac 1{2n}} \bigg) \\ & = \frac 1e (n+1 - n) \\ & = \boxed{\dfrac 1e} \end{aligned}

That's all?

Brian Lie - 3 years, 2 months ago

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Sorry, I forgot about it.

Chew-Seong Cheong - 3 years, 2 months ago

Finally got it.

Chew-Seong Cheong - 3 years, 2 months ago

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Thank you for the wonderful solution by Stirling's formula!

Brian Lie - 3 years, 2 months ago

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