anonymous
  • anonymous
Prove that the sum of n harmonic numbers H_1 + H_2 + ... + H_n = (n+1)H_n - n
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
What do you define as a harmonic number? \(H_n= \frac{1}{n}\)?
anonymous
  • anonymous
H_n =1+ 1/2 + 1/3 + ... 1/n
anonymous
  • anonymous
H_n is the actual sum from k=1 to n of 1/k

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anonymous
  • anonymous
So: \[H_n=\sum_{1 \leq i \leq n}\frac{1}{i}\]
anonymous
  • anonymous
Have you read Concrete Mathematics?
anonymous
  • anonymous
No, I'm in a Discrete Mathematics course right now. This is one of the challenge homework problems in the book.
anonymous
  • anonymous
Concrete Mathematics has an entire chapter dedicated to summation. You could learn various great concepts. This is even one particular problem. I recommend the book to you. I could attempt to explain the method, if you would like. It is a weebit complex, however.
anonymous
  • anonymous
I guess you could try.. the course assumes the student has a math background up to Calculus I. So far I've completed three calculus courses, ordinary differential equations, and linear algebra. Not sure if this problem pulls material from these latter courses though. Even so I doubt I'll understand what you're about to explain lol
anonymous
  • anonymous
But basically, a summation of a summation, right? Although I don't think the way to approach that is basic at all haha
anonymous
  • anonymous
Yes, that is step one. It requires manipulating the sum from there.
anonymous
  • anonymous
\[\sum_{j=1}^{n} H_n = \sum_{j=1}^{n} \sum_{k=1}^{j} \frac{1}{k}\] Lol, what's next?
anonymous
  • anonymous
Wait, first. Do you know what the Iverson Bracket is?
anonymous
  • anonymous
idk if I even typed that right.. aside from indexing errors
anonymous
  • anonymous
Sorry, but no
anonymous
  • anonymous
It's okay. Not many people do. But it is a notation used in summation to make things easier. The idea is this: \[ [x]= \left\{ \begin{array}{c} 1 &\text{if } x \text{ is true},\\ 0 &\text{otherwise.} \end{array} \right. \]
anonymous
  • anonymous
Okay, so this is relevant because it allows summation to be manipulated easily. When we say, for example, \(\sum_{1 \leq i \leq n}i\), we can equivalently say (using this notation) \[\sum_{i} i [1\leq i \leq n]\] This works because we take \(i\) to be evaluated from \(-\infty\) to \(\infty\) and the Iverson Bracket is zero at all points which do not satisfy our condition. Make sense?
anonymous
  • anonymous
(Oh, by the way, \(\sum_{1 \leq i \leq n}i\) is the same as \(\sum_{i=1}^{n}i\).)
anonymous
  • anonymous
actually.. we may actually be able to prove it without using that. the chapter I'm working on is about proofs by induction. a simple algebraic manipulation may work
anonymous
  • anonymous
Ah, yes. Induction is much easier. But this makes sense of _why_ it is how it is.
anonymous
  • anonymous
For induction, check \(n=1\). Then, assume for some integer \(k\), that the statement is true. Show that by adding \(H_{k+1}\) to the left side results in the right side.
anonymous
  • anonymous
Haha thanks.. Sorry for the trouble. Maybe I should've mentioned induction earlier
anonymous
  • anonymous
No, it's fine. You should learn about Iverson Brackets and etc anyway. Trust me, it makes almost every summation ridiculously easier.
anonymous
  • anonymous
Thanks, that might help in later computer science classes. The course I'm taking right now is like discrete math for comp sci
anonymous
  • anonymous
Concrete Mathematics is all about Discrete Mathematics. Take a look here on Wiki about it: http://en.wikipedia.org/wiki/Concrete_Mathematics
anonymous
  • anonymous
Oh wow, then I might really come across it in my later years of study. Thank you very much for the info!
anonymous
  • anonymous
You're welcome. :D

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