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Cubed Harmonic Sum

Prove That

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

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

Note by Ishan Singh
9 months, 2 weeks ago

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

Now, we can re-write it as: \[S=\left( \sum _{ { m }_{ 1 }=1 }^{ n }{ \frac { 1 }{ { m }_{ 1 } } } \right) \left( \sum _{ { m }_{ 2 }=1 }^{ n }{ \frac { 1 }{ { m }_{ 2 } } } \right) \left( \sum _{ { m }_{ 3 }=1 }^{ n }{ \frac { 1 }{ { m }_{ 3 } } } \right) \]

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

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

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

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

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

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: \[ \boxed{\displaystyle A=\sum _{ n=1 }^{ \infty }{ { \left( \frac { { H }_{ n } }{ n } \right) }^{ 3 } } =\frac { 31 }{ 5040 } { \pi }^{ 6 }-\frac { 5 }{ 2 } { \left( \zeta \left( 3 \right) \right) }^{ 2 }}\] Aditya Kumar · 9 months ago

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@Aditya Kumar Evaluate \(\zeta(3,2,1) \) with a proper solution. ?????? Thanks Aman Rajput · 9 months ago

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@Aman Rajput I found it in a website which I don't remember. It was a kind of blog website. Aditya Kumar · 9 months ago

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@Aditya Kumar which leads to an incomplete solution . try to find that blog Aman Rajput · 9 months ago

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@Aditya Kumar Nice solution! Ishan Singh · 9 months ago

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@Ishan Singh Thanks to you. Because of this I could learn many new concepts. Aditya Kumar · 9 months ago

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@Aditya Kumar Quasi - Shuffling can also be used to solve this question Ishan Singh · 9 months ago

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@Ishan Singh I'll try that tonight! Aditya Kumar · 9 months ago

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

\[\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^3 = \sum_{n=1}^{\infty} \left(\sum_{k=1}^{\infty} \frac{1}{k(k+n)}\right)^3\]\[= \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \sum_{k=1}^{\infty} \sum_{j=1}^{\infty} \frac{1}{jkm(j+n)(k+n)(m+n)}\] Ariel Gershon · 9 months, 1 week ago

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@Ariel Gershon Use \(\displaystyle H_{n} = \int_{0}^{1} \dfrac{1-x^n}{1-x} \mathrm{d}x \) and interchange sum and integral. Then try using Integration By Parts. Ishan Singh · 9 months, 1 week ago

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@Ishan Singh 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 · 9 months ago

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@Aditya Kumar \(H_{n} ^3 = \displaystyle \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \dfrac{(1-x^n)(1-y^n)(1-z^n)}{(1-x)(1-y)(1-z)} \mathrm{d}x \ \mathrm{d}y \ \mathrm{d}z\) Ishan Singh · 9 months ago

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@Ishan Singh Nice, I'll try this one also. Aditya Kumar · 9 months ago

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