Άθροισμα Euler με ουρά ζ(2)

Tolaso J Kos · #1

Να δείξετε ότι: $\displaystyle{\sum\limits_{n=1}^{\infty} \frac{\mathcal{H}_n}{n}\left(\zeta(2) - \sum\limits_{k=1}^{n} \frac{1}{n^2}\right) = \dfrac{7}{4}\zeta(4)}$ .

galactus · #2

Ξεκινάμε με τη γνωστή γεννήτρια:

$\displaystyle{1/2\left[\zeta(s)-\zeta(2s)\right]=\frac{(-1)^{s-1}}{(s-1)!}\int_{0}^{1}\frac{\log^{s-1}(v)Li_{s}(v)}{1-v}dv}$

s=2

:

$\displaystyle{\frac{-3}{4}\zeta(4)=\int_{0}^{1}\frac{\log(v)Li_{2}(v)}{1-v}dv.....[1]}$

$\displaystyle{\zeta(2)-\sum_{k=1}^{n}\frac{1}{k^{2}}=\zeta(2)-H_{n}^{(2)}=\psi_{1}(n+1)....[2]}$

$\displaystyle{\int_{0}^{1}u^{n-1}\log(1-u)du=\frac{H_{n}}{n}....[3]}$

$\displaystyle{\sum_{k=1}^{\infty}\int_{0}^{1}v^{n+k-1}\log(v)dy=\psi_{1}(n+1)......[4]}$

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σκεφτείτε:

$\displaystyle{\sum_{n=1}^{\infty}\frac{H_{n}}{n}\psi_{1}(n+1)}$

[3] και [4]:

$\displaystyle{\int_{0}^{1}u^{n-1}\log(1-u)du\sum_{k=1}^{\infty}\int_{0}^{1}v^{n+k-1}\log(v)dv}$

γεωμετρική σειρά:

$\displaystyle{\int_{0}^{1}\int_{0}^{1}u^{n-1}v^{n}\cdot \frac{\log(1-u)\log(v)}{1-v}dudv}$

έτσι:

$\displaystyle{\sum_{n=1}^{\infty}\frac{H_{n}}{n}\psi_{1}(n+1)=\sum_{n=1}^{\infty}\int_{0}^{1}\int_{0}^{1}\frac{u^{n-1}v^{n}\log(1-u)\log(v)}{1-v}dudv}$

$\displaystyle{=\int_{0}^{1}\frac{v\log(v)}{1-v}\int_{0}^{1}\frac{\log(1-v)}{1-uv}dudv}$

αλλά: $\displaystyle{\int_{0}^{1}\frac{\log(1-v)}{1-uv}du=\frac{1}{v}Li_{2}\left(\frac{v}{v-1}\right)}$

και $\displaystyle{-Li_{2}\left(\frac{v}{v-1}\right)=1/2\log^{2}(1-v)+Li_{2}(v)}$

όθεν $\displaystyle{-\int_{0}^{1}\frac{\log(v)}{1-v}\left(1/2\log^{2}(1-v)+Li_{2}(v)\right)=-1/2\int_{0}^{1}\frac{\log(v)\log^{2}(1-v)}{1-v}dv+\int_{0}^{1}\frac{\log(v)Li_{2}(v)}{1-v}dv}$

$\displaystyle{-1/2\cdot 2\left[1/2\log^{2}(1-v)Li_{2}(1-v)-\log(1-v)Li_{3}(1-v)+Li_{4}(1-v)-\zeta(4)\right]|_{0}^{1}+[1]}$

$\displaystyle{-\left(-\zeta(2)-\frac{3}{4}\zeta(4)\right)=\boxed{\frac{7}{4}\zeta(4)}}$

pprime · #3

Tolaso J Kos έγραψε:Να δείξετε ότι: $\displaystyle{\sum\limits_{n=1}^{\infty} \frac{\mathcal{H}_n}{n}\left(\zeta(2) - \sum\limits_{k=1}^{n} \frac{1}{n^2}\right) = \dfrac{7}{4}\zeta(4)}$ .

$\displaystyle{\sum\limits_{n=1}^{+\infty }{\frac{H_{n}}{n}\left( \zeta \left( 2 \right)-\sum\limits_{k=1}^{n}{\frac{1}{k^{2}}} \right)}=\sum\limits_{n=1}^{+\infty }{\frac{\psi _{1}\left( n+1 \right)}{n}H_{n}}=-\sum\limits_{n=1}^{+\infty }{\frac{H_{n}}{n}\int\limits_{0}^{1}{\frac{x^{n}\ln x}{1-x}dx}}=-\int\limits_{0}^{1}{\left( \frac{x\ln x}{1-x}\sum\limits_{n=1}^{+\infty }{\frac{H_{n}}{n}x^{n-1}} \right)dx}}$

$\displaystyle{=-\int\limits_{0}^{1}{\frac{x\ln x}{1-x}\left( \sum\limits_{n=0}^{+\infty }{\frac{x^{n}}{n+1}H_{n+1}} \right)dx}=-\int\limits_{0}^{1}{\frac{x\ln x}{1-x}\left( \frac{1}{x}\sum\limits_{n=0}^{+\infty }{\frac{H_{n}+\frac{1}{n+1}}{n+1}x^{n+1}} \right)dx}}$

$\displaystyle{=-\int\limits_{0}^{1}{\frac{x\ln x}{1-x}\left( 1+\frac{1}{x}\sum\limits_{n=1}^{+\infty }{\left( \frac{1}{\left( n+1 \right)^{2}}+\frac{H_{n}}{n+1} \right)x^{n+1}} \right)dx}}$

$\displaystyle{=-\int\limits_{0}^{1}{\frac{x\ln x}{1-x}\left( 1+\frac{1}{x}\sum\limits_{n=2}^{+\infty }{\frac{x^{n}}{n^{2}}}+\frac{1}{x}\sum\limits_{n=1}^{+\infty }{\frac{H_{n}}{n+1}x^{n+1}} \right)dx}=-\int\limits_{0}^{1}{\frac{\ln x}{1-x}\left( \frac{\ln ^{2}\left( 1-x \right)}{2}+\text{Li}_{2}\left( x \right) \right)dx}}$

$\displaystyle{=-\frac{1}{2}\int\limits_{0}^{1}{\frac{\ln x\ln ^{2}\left( 1-x \right)}{1-x}dx}-\int\limits_{0}^{1}{\frac{\ln x\text{Li}_{2}\left( x \right)}{1-x}dx}=\frac{\pi ^{4}}{90}+\frac{3}{4}\cdot \frac{\pi ^{4}}{90}=\frac{7\pi ^{4}}{360}}$

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(*)
$\displaystyle{\int\limits_{0}^{1}{t^{x-1}\left( 1-t \right)^{y-1}\ln t\ln ^{2}\left( 1-t \right)dt}=\frac{d^{2}}{dy^{2}}\left( \frac{d\beta \left( x,y \right)}{dx} \right)=\frac{d^{2}}{dy^{2}}\left( \left( \psi \left( x \right)-\psi \left( x+y \right) \right)\beta \left( x,y \right) \right)}$

$\displaystyle{=\left( \psi \left( x \right)-\psi \left( x+y \right) \right)\left( \left( \psi \left( y \right)-\psi \left( x+y \right) \right)^{2}\beta \left( x,y \right)+\left( \psi _{1}\left( y \right)-\psi _{1}\left( x+y \right) \right)\beta \left( x,y \right) \right)}$

$\displaystyle{-2\left( \psi \left( y \right)-\psi \left( x+y \right) \right)\psi _{1}\left( x+y \right)\beta \left( x,y \right)-\psi _{2}\left( x+y \right)\beta \left( x,y \right)}$

(**)
$\displaystyle{\int\limits_{0}^{1}{\frac{\ln x\text{Li}_{2}\left( x \right)}{1-x}dx}=\int\limits_{0}^{1}{\frac{\ln \left( 1-x \right)\text{Li}_{2}\left( 1-x \right)}{x}dx}}$

$\displaystyle{\text{Li}_{2}\left( x \right)+\text{Li}_{2}\left( 1-x \right)=\zeta \left( 2 \right)-\ln x\ln \left( 1-x \right)}$

$\displaystyle{\int\limits_{0}^{1}{\frac{\ln x\text{Li}_{2}\left( x \right)}{1-x}dx}=\zeta \left( 2 \right)\int\limits_{0}^{1}{\frac{\ln \left( 1-x \right)}{x}dx}-\int\limits_{0}^{1}{\frac{\ln \left( 1-x \right)\text{Li}_{2}\left( x \right)}{x}dx}-\int\limits_{0}^{1}{\frac{\ln \left( x \right)\ln ^{2}\left( 1-x \right)}{x}dx}}$

$\displaystyle{=-\zeta \left( 2 \right)\text{Li}_{2}\left( 1 \right)+\frac{1}{2}\left( \text{Li}_{2}\left( 1 \right) \right)^{2}-\int\limits_{0}^{1}{\frac{\ln \left( x \right)\ln ^{2}\left( 1-x \right)}{x}dx}=-\frac{3}{2}\left( \text{Li}_{2}\left( 1 \right) \right)^{2}-\frac{d^{2}}{dy^{2}}\left( \frac{d\beta \left( x,y \right)}{dx} \right)_{\left( x\to 0,y\to 1 \right)}}$

$\displaystyle{=\frac{1}{2}\left( \frac{\pi ^{2}}{6} \right)^{2}-\left( -\frac{\pi ^{4}}{180} \right)=\frac{\pi ^{4}}{180}-\frac{\pi ^{4}}{72}=-\frac{\pi ^{4}}{120}}$

Άθροισμα Euler με ουρά ζ(2)

Άθροισμα Euler με ουρά ζ(2)

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