Wow! (Who's up to the challenge? 11)

Calculus Level 5

∑ n = 1 ∞ ζ ( 2 n ) n ( 2 n + 1 ) 2 2 n = A ln ⁡ ( B π ) − C \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { \zeta (2n) }{ n(2n+1){ 2 }^{ 2n } } } =A\ln(B\pi )-C

where A , B , C A,B,C are positive integers, then find A + B + C A+B+C .

This is a part of Who's up to the challenge? .


The answer is 3.

Hamza A by Hamza A , 19, USA

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1 solution

Wesley Low
Aug 2, 2020

We first change the order of summation and rearrange the expression by partial fractions ∑ n = 1 ∞ ζ ( 2 n ) n ( 2 n + 1 ) 2 2 n = ∑ n = 1 ∞ ( 1 n − 2 2 n + 1 ) ∑ k = 1 ∞ 1 ( 2 k ) 2 n \sum_{n=1}^{\infty}\dfrac{\zeta\left(2n\right)}{n\left(2n+1\right)2^{2n}}=\sum_{n=1}^{\infty}\left(\dfrac{1}{n}-\dfrac{2}{2n+1}\right)\sum_{k=1}^{\infty}\dfrac{1}{\left(2k\right)^{2n}} = ∑ k = 1 ∞ ∑ n = 1 ∞ ( ( 1 4 k 2 ) n n − 4 k ( 1 2 k ) 2 n + 1 2 n + 1 ) =\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\left(\dfrac{\left(\dfrac{1}{4k^2}\right)^n}{n}-4k\dfrac{\left(\dfrac{1}{2k}\right)^{2n+1}}{2n+1}\right) Using the MacLaurin series for − ln ⁡ ( 1 − x ) -\ln\left(1-x\right) , we get ∑ k = 1 ∞ ( ∑ n = 1 ∞ ( 1 4 k 2 ) n n − 4 k ( ∑ n = 1 ∞ ( 1 2 k ) n n − ∑ n = 1 ∞ ( 1 2 k ) 2 n 2 n − ( 1 2 k ) 1 1 ) ) \sum_{k=1}^{\infty}\left(\sum_{n=1}^{\infty}\dfrac{\left(\dfrac{1}{4k^2}\right)^n}{n}-4k\left(\sum_{n=1}^{\infty}\dfrac{\left(\dfrac{1}{2k}\right)^{n}}{n}-\sum_{n=1}^{\infty}\dfrac{\left(\dfrac{1}{2k}\right)^{2n}}{2n}-\dfrac{\left(\dfrac{1}{2k}\right)^{1}}{1}\right)\right) = ∑ k = 1 ∞ ( − ln ⁡ ( 1 − 1 4 k 2 ) − 4 k ( − ln ⁡ ( 1 − 1 2 k ) + 1 2 ln ⁡ ( 1 − ( 1 2 k ) 2 ) − 1 2 k ) ) =\sum_{k=1}^{\infty}\left(-\ln\left(1-\dfrac{1}{4k^2}\right)-4k\left(-\ln\left(1-\dfrac{1}{2k}\right)+\dfrac{1}{2}\ln\left(1-\left(\dfrac{1}{2k}\right)^{2}\right)-\dfrac{1}{2k}\right)\right) = ∑ k = 1 ∞ ( − ( ln ⁡ ( 2 k + 1 ) + ln ⁡ ( 2 k − 1 ) − 2 ln ⁡ ( 2 k ) ) − 4 k ( − ln ⁡ ( 2 k − 1 ) + ln ⁡ ( 2 k ) + 1 2 ( ln ⁡ ( 2 k + 1 ) + ln ⁡ ( 2 k − 1 ) − 2 ln ⁡ ( 2 k ) ) ) + 2 ) =\sum_{k=1}^{\infty}\left(-\left(\ln\left(2k+1\right)+\ln\left(2k-1\right)-2\ln\left(2k\right)\right)-4k\left(-\ln\left(2k-1\right)+\ln\left(2k\right)+\dfrac{1}{2}\left(\ln\left(2k+1\right)+\ln\left(2k-1\right)-2\ln\left(2k\right)\right)\right)+2\right) = ∑ k = 1 ∞ ( 2 ln ⁡ ( 2 k ) − ( ( 2 k + 1 ) ln ⁡ ( 2 k + 1 ) − ( 2 k + 1 ) − ( ( 2 k − 1 ) ln ⁡ ( 2 k − 1 ) − ( 2 k − 1 ) ) ) ) =\sum_{k=1}^{\infty}\left(2\ln\left(2k\right)-\left(\left(2k+1\right)\ln\left(2k+1\right)-\left(2k+1\right)-\left(\left(2k-1\right)\ln\left(2k-1\right)-\left(2k-1\right)\right)\right)\right) Notice that this is actually the definite integral of ln ⁡ x \ln x , i.e. ∫ a b ln ⁡ x d x = b ln ⁡ b − b − ( a ln ⁡ a − a ) \int_{a}^{b}\ln xdx=b\ln b-b-\left(a\ln a-a\right) . Substituting this in gives us ∑ k = 1 ∞ ( 2 ln ⁡ ( 2 k ) − ∫ 2 k − 1 2 k + 1 ln ⁡ t d t ) \sum_{k=1}^{\infty}\left(2\ln\left(2k\right)-\int_{2k-1}^{2k+1}\ln t dt\right) = lim ⁡ n → ∞ ( 2 ∑ k = 1 n ln ⁡ ( 2 k ) − ∫ 1 2 n + 1 ln ⁡ t d t ) =\lim_{n\rightarrow\infty}\left(2\sum_{k=1}^{n}\ln\left(2k\right)-\int_{1}^{2n+1}\ln t dt\right) = lim ⁡ n → ∞ ( 2 n ln ⁡ 2 + 2 ln ⁡ n ! − ( 2 n + 1 ) ln ⁡ ( 2 n + 1 ) + ( 2 n + 1 ) − 1 ) =\lim_{n\rightarrow\infty}\left(2n\ln2+2\ln{n!}-\left(2n+1\right)\ln\left(2n+1\right)+\left(2n+1\right)-1\right) = lim ⁡ n → ∞ ln ⁡ ( ( n ! ) 2 ( 2 e ) 2 n ( 2 n + 1 ) 2 n + 1 ) =\lim_{n\rightarrow\infty}\ln\left(\dfrac{\left(n!\right)^2\left(2e\right)^{2n}}{\left(2n+1\right)^{2n+1}}\right) Since we are taking the limit as n → ∞ n\rightarrow\infty , we can apply Stirling's approximation for n ! n! , where n ! ∼ 2 π n ( n e ) n {n!}\sim\sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n . This gives lim ⁡ n → ∞ ln ⁡ ( ( 2 π n ( n e ) n ) 2 ( 2 e ) 2 n ( 2 n + 1 ) 2 n + 1 ) \lim_{n\rightarrow\infty}\ln\left(\dfrac{\left(\sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n\right)^2\left(2e\right)^{2n}}{\left(2n+1\right)^{2n+1}}\right) = lim ⁡ n → ∞ ln ⁡ ( π ( 2 n ) 2 n + 1 ( 2 n + 1 ) 2 n + 1 ) =\lim_{n\rightarrow\infty}\ln\left(\dfrac{\pi \left(2n\right)^{2n+1}}{\left(2n+1\right)^{2n+1}}\right) = lim ⁡ n → ∞ ln ⁡ ( π ( 1 + 1 2 n ) 2 n + 1 ) =\lim_{n\rightarrow\infty}\ln\left(\dfrac{\pi}{\left(1+\dfrac{1}{2n}\right)^{2n+1}}\right) = ln ⁡ π − lim ⁡ n → ∞ ln ⁡ ( ( 1 + 1 2 n ) 2 n + 1 ) =\ln\pi-\lim_{n\rightarrow\infty}\ln\left(\left(1+\dfrac{1}{2n}\right)^{2n+1}\right) = ln ⁡ π − lim ⁡ n → ∞ ln ⁡ ( ( 1 + 1 2 n ) 2 n ) − ln ⁡ ( ( 1 + 1 2 n ) ) =\ln\pi-\lim_{n\rightarrow\infty}\ln\left(\left(1+\dfrac{1}{2n}\right)^{2n}\right)-\ln\left(\left(1+\dfrac{1}{2n}\right)\right) = ln ⁡ π − ln ⁡ e − 0 =\ln\pi-\ln e - 0 = ln ⁡ π − 1 =\boxed{\ln\pi-1}

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