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
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Ishan Singh
2 years, 7 months ago
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Top NewestI 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.
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Yes even I thought about the same method! Shuffling indeed helps.
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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:
\[\boxed{S=\frac { { \zeta }^{ 2 }\left( m \right) +\zeta \left( 2m \right) }{ 2 } }\]
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Please correct me if I'm wrong.
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(+1) Correct!
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What was your method?
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What do you mean by 'summation by parts'?
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It is analogous to "Integration By Parts".
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I now one value of this.
Which is the first one.
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Nice note!
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