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

#Calculus

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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 - 4 years, 8 months ago

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

Aditya Kumar - 4 years, 8 months ago

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

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 - 4 years, 8 months ago

Please correct me if I'm wrong.

Aditya Kumar - 4 years, 8 months ago

(+1) Correct!

Ishan Singh - 4 years, 8 months ago

@Ishan Singh – What was your method?

Aditya Kumar - 4 years, 8 months ago

What do you mean by 'summation by parts'?

Deeparaj Bhat - 4 years, 8 months ago

It is analogous to "Integration By Parts".

Ishan Singh - 4 years, 8 months ago

I now one value of this.

Which is the first one.

Joel Yip - 4 years, 8 months ago

Nice note!

Joel Yip - 4 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$