Cool Summation!! (Hint: Fundamental something something)

Calculus Level 4

lim n [ ln ( 4 n 2 n ) + ln ( 16 n 2 n ) + ln ( 36 n 2 n ) + + ln ( 4 n 2 n 2 n ) ] = ? \lim_{n \to \infty} \left[\ln \left(\sqrt[n]{\frac{4}{n^2}}\right) + \ln\left(\sqrt[n]{\frac{16}{n^2}}\right) + \ln\left(\sqrt[n]{\frac{36}{n^2}}\right)+ \cdots + \ln\left(\sqrt[n]{\frac{4n^2}{n^2}}\right) \right]=\ ?

4 ln ( 2 ) 4\ln(2) 2 ln ( 4 ) 2 2\ln(4)-2 2 ln ( 2 ) 2 ln ( 2 ) 4 2\ln(2)-2\ln(2)-4 2 ln ( 2 ) 4 ln ( 4 ) 4 2\ln(2)-4\ln(4)-4 2 ln ( 2 ) 2 2\ln(2)-2 ln ( 4 ) 4 \ln(4)-4

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4 solutions

Chew-Seong Cheong
Mar 31, 2018

Relevant wiki: Riemann Sums

L = lim n [ ln ( 4 n 2 n ) + ln ( 16 n 2 n ) + ln ( 36 n 2 n ) + + ln ( 4 n 2 n 2 n ) ] = lim n k = 1 n ln ( 4 k 2 n 2 n ) = lim n k = 1 n 1 n ln ( 4 k 2 n 2 ) By Riemann sums: = 0 1 ln ( 4 x 2 ) d x lim n k = a b 1 n f ( k n ) = lim k n a k b k f ( x ) d x = 0 1 ( 2 ln 2 + 2 ln x ) d x = 2 x ln 2 + 2 x ln x 2 x 0 1 = 2 ln 2 + 2 ln 1 2 x 0 2 lim x 0 x ln x 0 See note: lim x 0 x ln x = 0 = 2 ln 2 2 \begin{aligned} L & = \lim_{n \to \infty} \left[\ln \left(\sqrt [n]{\frac 4{n^2}}\right) + \ln \left(\sqrt [n]{\frac {16}{n^2}}\right) + \ln \left(\sqrt [n]{\frac {36}{n^2}}\right) + \cdots + \ln \left(\sqrt [n]{\frac {4n^2}{n^2}}\right) \right] \\ & = \lim_{n \to \infty} \sum_{k=1}^n \ln \left(\sqrt [n]{\frac {4k^2}{n^2}}\right) \\ & = \lim_{n \to \infty} \sum_{k=1}^n \frac 1n \ln \left(\frac {4k^2}{n^2}\right) & \small \color{#3D99F6} \text{By Riemann sums: } \\ & = \int_0^1 \ln (4x^2)\ dx & \small \color{#3D99F6} \lim_{n \to \infty} \sum_{k=a}^b \frac 1n f\left(\frac kn\right) = \lim_{k \to n} \int_\frac ak^\frac bk f(x) \ dx \\ & = \int_0^1 \left(2\ln 2 + 2 \ln x\right)\ dx \\ & = 2x\ln 2 + 2x\ln x -2x\ \bigg|_0^1 \\ & = 2\ln 2 + 2\ln 1 - 2x - 0 - 2 {\color{#3D99F6} \lim_{x \to 0} x \ln x} - 0 & \small \color{#3D99F6} \text{See note: }\lim_{x \to 0} x \ln x = 0 \\ & = \boxed{2\ln 2 - 2} \end{aligned}


Note:

L = lim x 0 x ln x = lim x 0 ln x 1 x A / cases, L’H o ˆ pital’s rule applies. = lim x 0 1 x 1 x 2 Differentiate up and down w.r.t x . = lim x 0 x = 0 \begin{aligned} L & = \lim_{x \to 0} x\ln x \\ & = \lim_{x \to 0} \frac {\ln x}{\frac 1x} & \small \color{#3D99F6} \text{A }\infty/\infty \text{ cases, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \frac {\frac 1x}{-\frac 1{x^2}} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t }x. \\ & = \lim_{x \to 0} -x \\ & = 0 \end{aligned}

In the last step, i think you should limit it first because ln(x) tends to -infinite as x tends to 0

Lingga Musroji - 3 years, 2 months ago

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lim x 0 x ln x = lim x 0 ln x 1 x = 1 x 1 x 2 = 0 \lim_{x \to 0}x \ln x = \lim_{x \to 0} \frac {\ln x}{\frac 1x} = \frac {\frac 1x}{-\frac 1{x^2}} = 0 by L'Hopital rule.

Chew-Seong Cheong - 3 years, 2 months ago

I think you can use Stirling's approximation you get

L = lim n k = 1 n ln ( 4 k 2 n 2 n ) = lim n ln ( k = 1 n 4 k 2 n 2 n n ) = lim n ln ( 4 n 2 k = 1 n k 2 n ) = lim n ln ( 4 n 2 n ! n ! n ) = lim n ln ( 4 n 2 n 2 e 2 2 π n n ) = ln ( 4 e 2 1 1 ) = 2 ln ( 2 ) 2 \begin{aligned} L&=\lim_{n\to\infty}\sum\limits_{k=1}^n\ln\left(\sqrt[n]{\frac{ 4k^2}{n^2}}\right)=\lim_{n\to\infty}\ln\left(\sqrt[n]{\frac{\prod\limits_{k=1}^n 4k^2}{n^{2n}}}\right)=\lim_{n\to\infty}\ln\left(\frac{4}{n^2}\sqrt[n]{\prod\limits_{k=1}^n k^2}\right)\\ &=\lim_{n\to\infty}\ln\left(\frac{4}{n^2}\sqrt[n]{n!n!}\right)\\ &=\lim_{n\to\infty}\ln\left(\frac{4}{n^2}\frac{n^2}{e^2}\sqrt[n]{2\pi n}\right)=\ln\left(\frac{4}{e^2}\cdot 1\cdot 1\right)\\ &=2\ln(2)-2 \end{aligned}

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edit: If you think, it is too inexact because we use a approximation: No, because n n\to\infty :

More exactly you have: n ! = 2 π n ( n e ) n e μ ( n ) n!=\sqrt{2\pi n}\left(\frac n{\mathrm e}\right)^n\mathrm e^{\mu(n)} with 0 < μ ( x ) < 1 / ( 12 x ) , x > 0 lim n μ ( n ) = 0 0<\mu(x)<1/(12x)\,,\,x>0~~\Rightarrow \lim\limits_{n\to\infty}\mu(n)=0 , and so:

L 2 = lim n ln ( n ! n ! n ) = lim n ln ( n 2 e 2 2 π n n ) + lim n ln ( e 2 μ / n ) = lim n ln ( n 2 e 2 2 π n n ) \begin{aligned} L_2&=\lim_{n\to\infty}\ln\left(\sqrt[n]{n!n!}\right)=\lim_{n\to\infty}\ln\left(\frac{n^2}{e^2}\sqrt[n]{2\pi n}\right)+\lim_{n\to\infty}\ln\left(\mathrm e^{2\mu/n}\right)\\ &=\lim_{n\to\infty}\ln\left(\frac{n^2}{e^2}\sqrt[n]{2\pi n}\right) \end{aligned}

I think of Riemann sum and this one too but unable to use Stirling approximation . Thanks !!

Naren Bhandari - 3 years, 1 month ago

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You're welcome.

I have added something for better understanding, btw.

Florian Wechslberger - 3 years, 1 month ago
John Ross
Apr 1, 2018

Rearranging terms and using the log identity ln(a)+ln(b)+...ln(z) = ln(ab...z) gives us lim n ( 2 n ln ( 2 n n ! n n ) ) \lim_{n \to \infty} (\frac{2}{n} \ln(\frac{2^n n!}{n^n})) Using Stirling's approximation for n! ( 2 π n n + 1 2 e n n ! e n n + 1 2 e n \sqrt{2\pi}\ n^{n+{\small\frac12}}e^{-n} \le n! \le e\ n^{n+{\small\frac12}}e^{-n} ), we get lim n 2 ln ( 2 e k n n ) \lim_{n \to \infty} 2 \ln\big(\frac{2}{e} \sqrt[n]{k\sqrt{n}}\big) where 2 π k e \sqrt{2\pi}\ \le k \le e\ . This gives us the desired result 2 ln 2 2 2 \ln2-2

Timothy Cao
Mar 31, 2018

Amateur explanation:

This can be rewritten as a definite integral with bounds 0 to 2 as shown:

lim n i = 1 n l n ( ( 2 i n ) ( 2 n ) ) \lim_{n\to\infty} \sum_{i=1}^{n}ln((\frac{2i}{n})^{(\frac{2}{n})})

= lim n i = 1 n ( 2 n ) l n ( 2 i n ) =\lim_{n\to\infty} \sum_{i=1}^{n}(\frac{2}{n})ln(\frac{2i}{n})

= lim n ( 2 n ) i = 1 n l n ( 2 i n ) =\lim_{n\to\infty}(\frac{2}{n}) \sum_{i=1}^{n}ln(\frac{2i}{n})

= 0 2 l n ( x ) d x =\int_{0}^{2} ln(x) dx

= 2 l n ( 2 ) 2 =2ln(2)-2

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