Prove that the sum of n harmonic numbers H_1 + H_2 + ... + H_n = (n+1)H_n - n

- anonymous

Prove that the sum of n harmonic numbers H_1 + H_2 + ... + H_n = (n+1)H_n - n

- schrodinger

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- anonymous

What do you define as a harmonic number? \(H_n= \frac{1}{n}\)?

- anonymous

H_n =1+ 1/2 + 1/3 + ... 1/n

- anonymous

H_n is the actual sum from k=1 to n of 1/k

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## More answers

- anonymous

So: \[H_n=\sum_{1 \leq i \leq n}\frac{1}{i}\]

- anonymous

Have you read Concrete Mathematics?

- anonymous

No, I'm in a Discrete Mathematics course right now. This is one of the challenge homework problems in the book.

- 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

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

But basically, a summation of a summation, right? Although I don't think the way to approach that is basic at all haha

- anonymous

Yes, that is step one. It requires manipulating the sum from there.

- anonymous

\[\sum_{j=1}^{n} H_n = \sum_{j=1}^{n} \sum_{k=1}^{j} \frac{1}{k}\]
Lol, what's next?

- anonymous

Wait, first. Do you know what the Iverson Bracket is?

- anonymous

idk if I even typed that right.. aside from indexing errors

- anonymous

Sorry, but no

- 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

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

(Oh, by the way, \(\sum_{1 \leq i \leq n}i\) is the same as \(\sum_{i=1}^{n}i\).)

- 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

Ah, yes. Induction is much easier. But this makes sense of _why_ it is how it is.

- 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

Haha thanks.. Sorry for the trouble. Maybe I should've mentioned induction earlier

- anonymous

No, it's fine. You should learn about Iverson Brackets and etc anyway. Trust me, it makes almost every summation ridiculously easier.

- 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

Concrete Mathematics is all about Discrete Mathematics. Take a look here on Wiki about it: http://en.wikipedia.org/wiki/Concrete_Mathematics

- anonymous

Oh wow, then I might really come across it in my later years of study. Thank you very much for the info!

- anonymous

You're welcome. :D

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