Cubed Harmonic Sum

Prove That

∑n=1∞(Hnn)3=315040π6−52(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

Notations:

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

Note by Ishan Singh
5 years, 3 months ago

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Comments

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=(∑n≥m1>m2>m3>1+∑n≥m1>m3>m2>1+∑n≥m2>m3>m1>1+∑n≥m2>m1>m3>1+∑n≥m3>m2>m1>1+∑n≥m3>m1>m2>1+∑n≥m1>m3=m2>1+∑n≥m2>m3=m1>1+∑n≥m3>m1=m2>1+∑n≥m1=m3>m2>1+∑n≥m2=m3>m2>1+∑n≥m1=m2>m3>1+∑n≥m1=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=1∞6Hn(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=1∞Hn(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=315040π6−52(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 - 5 years, 3 months ago

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

Aman Rajput - 5 years, 3 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 - 5 years, 3 months ago

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

Aman Rajput - 5 years, 3 months ago

Nice solution!

Ishan Singh - 5 years, 3 months ago

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

Aditya Kumar - 5 years, 3 months ago

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

Ishan Singh - 5 years, 3 months ago

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@Ishan Singh – I'll try that tonight!

Aditya Kumar - 5 years, 3 months ago

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=1∞1k(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=1∞∑m=1∞∑k=1∞∑j=1∞1jkm(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 - 5 years, 3 months ago

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Use Hn=∫011−xn1−xdx\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 - 5 years, 3 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 - 5 years, 3 months ago

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@Aditya Kumar – Hn3=∫01∫01∫01(1−xn)(1−yn)(1−zn)(1−x)(1−y)(1−z)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 - 5 years, 3 months ago

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

Aditya Kumar - 5 years, 3 months ago
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