Enormous Harmonic Sum proof (help mehhh)

My friend asked me a question about this (from Russia IMO or something I forgot). The question asked for n=2013n=2013. I know the answer by just trying small numbers and then predict. Here's the question to prove:

Prove that j=1n(i=jn1i)2+i=1n1i=2n\sum\limits_{j=1}^n (\sum\limits_{i=j}^n \frac{1}{i})^{2} + \sum\limits_{i=1}^n \frac{1}{i} = 2n

If you're confused with these stuffs, it is...

(11+12+...+1n)2+(12+...+1n)2+.......+(1n)2+11+12+...+1n=2n(\frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n})^{2} + (\frac{1}{2} + ... + \frac{1}{n})^{2} + ....... + (\frac{1}{n})^{2} + \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} = 2n

Note by Samuraiwarm Tsunayoshi
6 years 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} ai1 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 sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

Let Hn=i=1n1i H_n = \sum_{i=1}^n \frac1{i} (with the convention that H0=0 H_0 = 0 ). My approach was to write the quantity on the left side as Sn=j=1n(HnHj1)2+Hn S_n = \sum_{j=1}^n (H_n - H_{j-1})^2 + H_n . It is enough to show that Sn+1Sn=2 S_{n+1}-S_n = 2 , since S1=2 S_1 = 2 is the base case for an easy induction.

This difference equals j=1n+1((Hn+1Hj1)2(HnHj1)2)+1n+1 \sum_{j=1}^{n+1} \left( (H_{n+1}-H_{j-1})^2-(H_n-H_{j-1})^2 \right) + \frac1{n+1} , which simplifies by a difference of squares formula to j=1n+11n+1(Hn+1+Hn2Hj1)+1n+1 \sum_{j=1}^{n+1} \frac1{n+1} ( H_{n+1} + H_n - 2 H_{j-1}) + \frac1{n+1} , which becomes Hn+1+Hn+1n+12n+1j=1n+1Hj1=2Hn+12n+1j=1n+1Hj1 H_{n+1} + H_n + \frac1{n+1} - \frac2{n+1} \sum_{j=1}^{n+1} H_{j-1} = 2 H_{n+1} - \frac2{n+1} \sum_{j=1}^{n+1} H_{j-1} .

Ah, but now we can use the identity i=0nHi=(n+1)Hnn \sum_{i=0}^n H_i = (n+1)H_n - n , which is easy to prove by induction. (I found it on the wikipedia page for "harmonic number.") Plugging this in, we get 2Hn+12n+1((n+1)Hnn)=2Hn+12Hn+2nn+1=2n+1+2nn+1=2 2 H_{n+1} - \frac2{n+1} ((n+1) H_n - n) = 2 H_{n+1} - 2 H_n + \frac{2n}{n+1} = \frac2{n+1} + \frac{2n}{n+1} = 2 as desired.

There is probably a much more elegant way to do the problem (i.e. directly without induction), but this certainly works.

Patrick Corn - 6 years ago

Log in to reply

I don't quite understand how it is enough to prove that Sn+1Sn=2S_{n+1} - S_{n} = 2. I understand the rest but I'm just stuck at the first part. =_=" Sorry about that *cries

Log in to reply

Sn+1Sn=2(n+1)2n=2S_{n+1} - S_{n} = 2(n+1) - 2n = 2

Ronald Overwater - 6 years ago

Log in to reply

@Ronald Overwater Oh, my bad. =__="

Log in to reply

Thank you for the proof. ^__^

Log in to reply

This is from IMC 2013. :)

Zi Song Yeoh - 6 years ago

Log in to reply

Thanks! ^__^

Log in to reply


Problem Loading...

Note Loading...

Set Loading...