Generalized Harmonic Summation

Evaluate

∑r=1∞Hr(m)rm;m≥2\large{\sum_{r=1}^{\infty} \dfrac{H_{r} ^{(m)}}{r^m}} \quad ; \quad m \geq 2

Notation: Hr(m)H_{r} ^{(m)} denotes the Generalized Harmonic Number.

This is a part of the set Formidable Series and Integrals

#Calculus

Note by Ishan Singh
3 years, 10 months ago

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I have solved it by 44 methods. One is Summation By Parts, other are the following :

First, expand the summation, using the definition of Hr(m)\displaystyle H_{r}^{(m)} , to see that ∑r=1nHr(m)rm=∑r=1n1r2m+∑r<j∑j=2n1rj=Hn(2m)+∑r<j∑j=2n1rj\displaystyle \sum_{r=1}^{n} \dfrac{H_{r}^{(m)}}{r^m} = \sum_{r=1}^{n} \dfrac{1}{r^{2m}} + \sum_{r < j} \sum_{j=2}^{n} \dfrac{1}{rj} = H_{n}^{(2m)} + \sum_{r < j} \sum_{j=2}^{n} \dfrac{1}{rj}

Also,

[Hn(m)]2=Hn(2m)+2∑r<j∑j=2n1rj\displaystyle [H_{n}^{(m)}]^2 = H_{n}^{(2m)} + 2\sum_{r < j} \sum_{j=2}^{n} \dfrac{1}{rj}

Eliminating ∑r<j∑j=2n1rj\displaystyle \sum_{r < j} \sum_{j=2}^{n} \dfrac{1}{rj} from the above equations, we have,

∑r=1nHr(m)rm=12([Hn(m)]2+Hn(2m))(*)\displaystyle \sum_{r=1}^{n} \dfrac{H_{r}^{(m)}}{r^m} =\dfrac{1}{2} \left( [H_{n}^{(m)}]^2 + H_{n}^{(2m)} \right) \tag{*}

Now take limit to infinity to get the desired result.

Also note that (∗)(*) holds for m=1m=1 as well (but the infinite sum diverges).

My other two methods use integral representations of Hr(m)\displaystyle H_{r}^{(m)} and 1rm\dfrac{1}{r^m}. For instance, I have used one of them here.

Ishan Singh - 3 years, 10 months ago

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Yes even I thought about the same method! Shuffling indeed helps.

Aditya Kumar - 3 years, 10 months ago

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S=lim⁡n→∞∑r=1nHr(m)rm\displaystyle S=\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { { H }_{ r }^{ \left( m \right) } }{ { r }^{ m } } } }

By Summation by parts, we get:

S=lim⁡n→∞Hn(m)Hn+1(m)−∑r=1nHr(m)(Hr+1(m)−Hr(m))\displaystyle S=\lim _{ n\rightarrow \infty }{ { H }_{ n }^{ \left( m \right) }{ H }_{ n+1 }^{ \left( m \right) }-\sum _{ r=1 }^{ n }{ { H }_{ r }^{ \left( m \right) }\left( { H }_{ r+1 }^{ \left( m \right) }-{ H }_{ r }^{ \left( m \right) } \right) } }

Now we use Hr+1(m)=Hr(m)+1(r+1)m{ H }_{ r+1 }^{ \left( m \right) }={ H }_{ r }^{ \left( m \right) }+\frac { 1 }{ { \left( r+1 \right) }^{ m } }

S=lim⁡n→∞Hn(m)Hn+1(m)−∑r=1nHr(m)(r+1)m\displaystyle S=\lim _{ n\rightarrow \infty }{ { H }_{ n }^{ \left( m \right) }{ H }_{ n+1 }^{ \left( m \right) }-\sum _{ r=1 }^{ n }{ \frac { { H }_{ r }^{ \left( m \right) } }{ { \left( r+1 \right) }^{ m } } } }

Again, we use Hr+1(m)=Hr(m)+1(r+1)m{ H }_{ r+1 }^{ \left( m \right) }={ H }_{ r }^{ \left( m \right) }+\frac { 1 }{ { \left( r+1 \right) }^{ m } }

S=lim⁡n→∞Hn(m)Hn+1(m)−S+1+∑r=1n1(r+1)2m\displaystyle S=\lim _{ n\rightarrow \infty }{ { H }_{ n }^{ \left( m \right) }{ H }_{ n+1 }^{ \left( m \right) }-S+1+\sum _{ r=1 }^{ n }{ \frac { 1 }{ { \left( r+1 \right) }^{ 2m } } } }

On simplifying, we get:

S=ζ2(m)+ζ(2m)2\boxed{S=\frac { { \zeta }^{ 2 }\left( m \right) +\zeta \left( 2m \right) }{ 2 } }

Aditya Kumar - 3 years, 10 months ago

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Please correct me if I'm wrong.

Aditya Kumar - 3 years, 10 months ago

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(+1) Correct!

Ishan Singh - 3 years, 10 months ago

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@Ishan Singh – What was your method?

Aditya Kumar - 3 years, 10 months ago

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What do you mean by 'summation by parts'?

Deeparaj Bhat - 3 years, 10 months ago

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It is analogous to "Integration By Parts".

Ishan Singh - 3 years, 10 months ago

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I now one value of this.

Which is the first one.

Joel Yip - 3 years, 10 months ago

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Nice note!

Joel Yip - 3 years, 10 months ago

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