Generalized Harmonic Summation

Evaluate

∑r=1∞Hr(m)rm;m≥2\large{\sum_{r=1}^{\infty} \dfrac{H_{r} ^{(m)}}{r^m}} \quad ; \quad m \geq 2

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

This is a part of the set Formidable Series and Integrals

#Calculus

Note by Ishan Singh
3 years, 7 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

I have solved it by 44 methods. One is Summation By Parts, other are the following :

First, expand the summation, using the definition of Hr(m)\displaystyle H_{r}^{(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}

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}

Eliminating ∑r<j∑j=2n1rj\displaystyle \sum_{r < j} \sum_{j=2}^{n} \dfrac{1}{rj} 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{*}

Now take limit to infinity to get the desired result.

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

My other two methods use integral representations of Hr(m)\displaystyle H_{r}^{(m)} and 1rm\dfrac{1}{r^m}. For instance, I have used one of them here.

Ishan Singh - 3 years, 7 months ago

Log in to reply

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

Aditya Kumar - 3 years, 7 months ago

Log in to reply

S=lim⁡n→∞∑r=1nHr(m)rm\displaystyle S=\lim _{ n\rightarrow \infty }{ \sum _{ r=1 }^{ n }{ \frac { { H }_{ r }^{ \left( m \right) } }{ { r }^{ m } } } }

By Summation by parts, we get:

S=lim⁡n→∞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) } }

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 } }

S=lim⁡n→∞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 } } } }

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 } }

S=lim⁡n→∞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 } } } }

On simplifying, we get:

S=ζ2(m)+ζ(2m)2\boxed{S=\frac { { \zeta }^{ 2 }\left( m \right) +\zeta \left( 2m \right) }{ 2 } }

Aditya Kumar - 3 years, 7 months ago

Log in to reply

Please correct me if I'm wrong.

Aditya Kumar - 3 years, 7 months ago

Log in to reply

(+1) Correct!

Ishan Singh - 3 years, 7 months ago

Log in to reply

@Ishan Singh – What was your method?

Aditya Kumar - 3 years, 7 months ago

Log in to reply

What do you mean by 'summation by parts'?

Deeparaj Bhat - 3 years, 7 months ago

Log in to reply

It is analogous to "Integration By Parts".

Ishan Singh - 3 years, 7 months ago

Log in to reply

I now one value of this.

Which is the first one.

Joel Yip - 3 years, 7 months ago

Log in to reply

Nice note!

Joel Yip - 3 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...