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Generalized Harmonic Summation

Evaluate

\[\large{\sum_{r=1}^{\infty} \dfrac{H_{r} ^{(m)}}{r^m}} \quad ; \quad m \geq 2\]

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


This is a part of the set Formidable Series and Integrals

Note by Ishan Singh
3 months, 3 weeks ago

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I have solved it by \(4\) methods. One is Summation By Parts, other are the following :

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

Also,

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

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

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

Now take limit to infinity to get the desired result.

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

My other two methods use integral representations of \(\displaystyle H_{r}^{(m)}\) and \(\dfrac{1}{r^m}\). For instance, I have used one of them here. Ishan Singh · 3 months, 3 weeks ago

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@Ishan Singh Yes even I thought about the same method! Shuffling indeed helps. Aditya Kumar · 3 months, 3 weeks ago

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\(\displaystyle S=\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { { H }_{ r }^{ \left( m \right) } }{ { r }^{ m } } } } \)

By Summation by parts, we get:

\(\displaystyle S=\lim _{ n\rightarrow \infty }{ { H }_{ n }^{ \left( m \right) }{ H }_{ n+1 }^{ \left( m \right) }-\sum _{ r=1 }^{ n }{ { H }_{ r }^{ \left( m \right) }\left( { H }_{ r+1 }^{ \left( m \right) }-{ H }_{ r }^{ \left( m \right) } \right) } } \)

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

\(\displaystyle S=\lim _{ n\rightarrow \infty }{ { H }_{ n }^{ \left( m \right) }{ H }_{ n+1 }^{ \left( m \right) }-\sum _{ r=1 }^{ n }{ \frac { { H }_{ r }^{ \left( m \right) } }{ { \left( r+1 \right) }^{ m } } } } \)

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

\(\displaystyle S=\lim _{ n\rightarrow \infty }{ { H }_{ n }^{ \left( m \right) }{ H }_{ n+1 }^{ \left( m \right) }-S+1+\sum _{ r=1 }^{ n }{ \frac { 1 }{ { \left( r+1 \right) }^{ 2m } } } } \)

On simplifying, we get:

\[\boxed{S=\frac { { \zeta }^{ 2 }\left( m \right) +\zeta \left( 2m \right) }{ 2 } }\] Aditya Kumar · 3 months, 3 weeks ago

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@Aditya Kumar What do you mean by 'summation by parts'? Deeparaj Bhat · 3 months, 2 weeks ago

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@Deeparaj Bhat It is analogous to "Integration By Parts". Ishan Singh · 3 months, 2 weeks ago

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@Aditya Kumar Please correct me if I'm wrong. Aditya Kumar · 3 months, 3 weeks ago

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@Aditya Kumar (+1) Correct! Ishan Singh · 3 months, 3 weeks ago

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@Ishan Singh What was your method? Aditya Kumar · 3 months, 3 weeks ago

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Nice note! Joel Yip · 3 months, 3 weeks ago

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I now one value of this.

Which is the first one. Joel Yip · 3 months, 3 weeks ago

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