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Evaluate ∑r=1∞Hr(m)rm;m≥2\large{\sum_{r=1}^{\infty} \dfrac{H_{r} ^{(m)}}{r^m}} \quad ; \quad m \geq 2r=1∑∞rmHr(m);m≥2
Evaluate
∑r=1∞Hr(m)rm;m≥2\large{\sum_{r=1}^{\infty} \dfrac{H_{r} ^{(m)}}{r^m}} \quad ; \quad m \geq 2r=1∑∞rmHr(m);m≥2
Notation: Hr(m)H_{r} ^{(m)}Hr(m) denotes the Generalized Harmonic Number.
This is a part of the set Formidable Series and Integrals
Note by Ishan Singh 4 years, 8 months ago
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I have solved it by 444 methods. One is Summation By Parts, other are the following :
First, expand the summation, using the definition of Hr(m)\displaystyle H_{r}^{(m)}Hr(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} r=1∑nrmHr(m)=r=1∑nr2m1+r<j∑j=2∑nrj1=Hn(2m)+r<j∑j=2∑nrj1
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} [Hn(m)]2=Hn(2m)+2r<j∑j=2∑nrj1
Eliminating ∑r<j∑j=2n1rj\displaystyle \sum_{r < j} \sum_{j=2}^{n} \dfrac{1}{rj}r<j∑j=2∑nrj1 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{*} r=1∑nrmHr(m)=21([Hn(m)]2+Hn(2m))(*)
Now take limit to infinity to get the desired result.
Also note that (∗)(*)(∗) holds for m=1m=1m=1 as well (but the infinite sum diverges).
My other two methods use integral representations of Hr(m)\displaystyle H_{r}^{(m)}Hr(m) and 1rm\dfrac{1}{r^m}rm1. For instance, I have used one of them here.
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Yes even I thought about the same method! Shuffling indeed helps.
S=limn→∞∑r=1nHr(m)rm\displaystyle S=\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { { H }_{ r }^{ \left( m \right) } }{ { r }^{ m } } } } S=n→∞limr=1∑nrmHr(m)
By Summation by parts, we get:
S=limn→∞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) } } S=n→∞limHn(m)Hn+1(m)−r=1∑nHr(m)(Hr+1(m)−Hr(m))
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 } } Hr+1(m)=Hr(m)+(r+1)m1
S=limn→∞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 } } } } S=n→∞limHn(m)Hn+1(m)−r=1∑n(r+1)mHr(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 } } Hr+1(m)=Hr(m)+(r+1)m1
S=limn→∞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 } } } } S=n→∞limHn(m)Hn+1(m)−S+1+r=1∑n(r+1)2m1
On simplifying, we get:
S=ζ2(m)+ζ(2m)2\boxed{S=\frac { { \zeta }^{ 2 }\left( m \right) +\zeta \left( 2m \right) }{ 2 } }S=2ζ2(m)+ζ(2m)
Please correct me if I'm wrong.
(+1) Correct!
@Ishan Singh – What was your method?
What do you mean by 'summation by parts'?
It is analogous to "Integration By Parts".
I now one value of this.
Which is the first one.
Nice note!
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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 Hr(m) , to see that r=1∑nrmHr(m)=r=1∑nr2m1+r<j∑j=2∑nrj1=Hn(2m)+r<j∑j=2∑nrj1
Also,
[Hn(m)]2=Hn(2m)+2r<j∑j=2∑nrj1
Eliminating r<j∑j=2∑nrj1 from the above equations, we have,
r=1∑nrmHr(m)=21([Hn(m)]2+Hn(2m))(*)
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 Hr(m) and rm1. 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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S=n→∞limr=1∑nrmHr(m)
By Summation by parts, we get:
S=n→∞limHn(m)Hn+1(m)−r=1∑nHr(m)(Hr+1(m)−Hr(m))
Now we use Hr+1(m)=Hr(m)+(r+1)m1
S=n→∞limHn(m)Hn+1(m)−r=1∑n(r+1)mHr(m)
Again, we use Hr+1(m)=Hr(m)+(r+1)m1
S=n→∞limHn(m)Hn+1(m)−S+1+r=1∑n(r+1)2m1
On simplifying, we get:
S=2ζ2(m)+ζ(2m)
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Please correct me if I'm wrong.
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(+1) Correct!
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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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