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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestI 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.

Log in to reply

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

Log in to reply

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

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:

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

Log in to reply

Please correct me if I'm wrong.

Log in to reply

(+1) Correct!

Log in to reply

Log in to reply

What do you mean by 'summation by parts'?

Log in to reply

It is analogous to "Integration By Parts".

Log in to reply

I now one value of this.

Which is the first one.

Log in to reply

Nice note!

Log in to reply