Cubed Harmonic Sum

Prove That

n=1(Hnn)3=315040π652(ζ(3))2\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^3=\dfrac{31}{5040}\pi^6-\dfrac{5}{2}(\zeta(3))^2

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This is a part of the set Formidable Series and Integrals.

Note by Ishan Singh
3 years, 7 months ago

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First I'll evaluate: S=(m=1n1m)3\displaystyle S={ \left( \sum _{ m=1 }^{ n }{ \frac { 1 }{ m } } \right) }^{ 3 }

Now, we can re-write it as: S=(m1=1n1m1)(m2=1n1m2)(m3=1n1m3)S=\left( \sum _{ { m }_{ 1 }=1 }^{ n }{ \frac { 1 }{ { m }_{ 1 } } } \right) \left( \sum _{ { m }_{ 2 }=1 }^{ n }{ \frac { 1 }{ { m }_{ 2 } } } \right) \left( \sum _{ { m }_{ 3 }=1 }^{ n }{ \frac { 1 }{ { m }_{ 3 } } } \right)

This satisfies quasi-shuffle identity. So I can re-write it as: S=(nm1>m2>m3>1+nm1>m3>m2>1+nm2>m3>m1>1+nm2>m1>m3>1+nm3>m2>m1>1+nm3>m1>m2>1+nm1>m3=m2>1+nm2>m3=m1>1+nm3>m1=m2>1+nm1=m3>m2>1+nm2=m3>m2>1+nm1=m2>m3>1+nm1=m2=m3>1)1m1m2m3S=\left( \sum _{ n\ge { m }_{ 1 }>{ m }_{ 2 }>{ m }_{ 3 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 1 }>{ m }_{ 3 }>{ m }_{ 2 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 2 }>{ m }_{ 3 }>{ m }_{ 1 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 2 }>{ m }_{ 1 }>{ m }_{ 3 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 3 }>{ m }_{ 2 }>{ m }_{ 1 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 3 }>{ m }_{ 1 }>{ m }_{ 2 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 1 }>{ m }_{ 3 }={ m }_{ 2 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 2 }>{ m }_{ 3 }={ m }_{ 1 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 3 }>{ m }_{ 1 }={ m }_{ 2 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 1 }={ m }_{ 3 }>{ m }_{ 2 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 2 }={ m }_{ 3 }>{ m }_{ 2 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 1 }={ m }_{ 2 }>{ m }_{ 3 }>1 }^{ }{ + } \sum _{ n\ge { m }_{ 1 }={ m }_{ 2 }={ m }_{ 3 }>1 }^{ }{ } \right) \frac { 1 }{ { m }_{ 1 }{ m }_{ 2 }{ m }_{ 3 } }

Now, using multi-harmonic sum I'll re-write it as: S=6Hn(1,1,1)+3Hn(2,1)+3Hn(1,2)+Hn(3)S={ 6H }_{ n }\left( 1,1,1 \right) +{ 3H }_{ n }\left( 2,1 \right) +{ 3H }_{ n }\left( 1,2 \right) +{ H }_{ n }\left( 3 \right)

Now, coming back to the problem. I'll insert the value of S here. A=n=16Hn(1,1,1)+3Hn(2,1)+3Hn(1,2)+Hn(3)n3A=\sum _{ n=1 }^{ \infty }{ \frac { { 6H }_{ n }\left( 1,1,1 \right) +{ 3H }_{ n }\left( 2,1 \right) +{ 3H }_{ n }\left( 1,2 \right) +{ H }_{ n }\left( 3 \right) }{ { n }^{ 3 } } }

Now, I'll use the relation of multi-harmonic sum and multi-zeta variable: n=1Hn(s1,,sk)ns=ζ(s,s1,,sk)+ζ(s+s1,s2,,sk).\sum_{n=1}^\infty \frac{H_n(s_1,\ldots,s_k)}{n^s}=\zeta(s,s_1,\ldots,s_k)+\zeta(s+s_1,s_2,\ldots,s_k).

Therefore, A=6ζ(3,1,1,1)+6ζ(4,1,1)+3ζ(3,2,1)+3ζ(5,1)+3ζ(3,1,1)+3ζ(4,1)+ζ(3,3)+ζ(6)A=6\zeta \left( 3,1,1,1 \right) +6\zeta \left( 4,1,1 \right) +3\zeta \left( 3,2,1 \right) +3\zeta \left( 5,1 \right) +3\zeta \left( 3,1,1 \right) +3\zeta \left( 4,1 \right) +\zeta \left( 3,3 \right) +\zeta \left( 6 \right)

Now, on inserting values of each zeta (took me days to get each one of them and each took pages to solve and hence I'm not posting the method), we get: A=n=1(Hnn)3=315040π652(ζ(3))2 \boxed{\displaystyle A=\sum _{ n=1 }^{ \infty }{ { \left( \frac { { H }_{ n } }{ n } \right) }^{ 3 } } =\frac { 31 }{ 5040 } { \pi }^{ 6 }-\frac { 5 }{ 2 } { \left( \zeta \left( 3 \right) \right) }^{ 2 }}

Aditya Kumar - 3 years, 7 months ago

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Evaluate ζ(3,2,1)\zeta(3,2,1) with a proper solution. ?????? Thanks

Aman Rajput - 3 years, 7 months ago

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I found it in a website which I don't remember. It was a kind of blog website.

Aditya Kumar - 3 years, 7 months ago

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@Aditya Kumar which leads to an incomplete solution . try to find that blog

Aman Rajput - 3 years, 7 months ago

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Nice solution!

Ishan Singh - 3 years, 7 months ago

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Thanks to you. Because of this I could learn many new concepts.

Aditya Kumar - 3 years, 7 months ago

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@Aditya Kumar Quasi - Shuffling can also be used to solve this question

Ishan Singh - 3 years, 7 months ago

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@Ishan Singh I'll try that tonight!

Aditya Kumar - 3 years, 7 months ago

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Yes, could you please give a hint? I'm stumped, this is as far as I could get:

n=1(Hnn)3=n=1(k=11k(k+n))3\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^3 = \sum_{n=1}^{\infty} \left(\sum_{k=1}^{\infty} \frac{1}{k(k+n)}\right)^3=n=1m=1k=1j=11jkm(j+n)(k+n)(m+n)= \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \sum_{k=1}^{\infty} \sum_{j=1}^{\infty} \frac{1}{jkm(j+n)(k+n)(m+n)}

Ariel Gershon - 3 years, 7 months ago

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Use Hn=011xn1xdx\displaystyle H_{n} = \int_{0}^{1} \dfrac{1-x^n}{1-x} \mathrm{d}x and interchange sum and integral. Then try using Integration By Parts.

Ishan Singh - 3 years, 7 months ago

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Are you sure about interchanging the integral and summation sigh? I don't think that will be possible as you are taking the integral inside the cube.

Aditya Kumar - 3 years, 7 months ago

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@Aditya Kumar Hn3=010101(1xn)(1yn)(1zn)(1x)(1y)(1z)dx dy dzH_{n} ^3 = \displaystyle \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \dfrac{(1-x^n)(1-y^n)(1-z^n)}{(1-x)(1-y)(1-z)} \mathrm{d}x \ \mathrm{d}y \ \mathrm{d}z

Ishan Singh - 3 years, 7 months ago

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@Ishan Singh Nice, I'll try this one also.

Aditya Kumar - 3 years, 7 months ago

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