# $n$'th Harmonic Number Proof

I couldn't find any simpler explanation of how to prove the n'th harmonic number, so here I am asking...

The $n$'th harmonic number is defined as $$H_{n} = \displaystyle\sum_{i=1}^n \frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$$ Show that $$H_n = \displaystyle \int_0^1 \frac{1 - x^n}{1 - x}\,dx$$

So I don't have a clue on how to approach this. I'm in a first year undergrad class and so far we have only learned about limits, derivatives and integrals. For integrals, we have only gone up to learning the substitution rule... So is there a way to approach and answer this question using only integrals (and limits?) without the use of series and sequences or any other sort of knowledge to that kind?

• Interesting username – user285523 Nov 30 '15 at 5:00

This becomes a straightforward integral when you use the identity $$\frac{1-x^n}{1-x}=1+x+x^2+\dots+x^{n-1}.$$ To prove this identity, note that $$(1+x+x^2+\dots+x^{n-1})(1-x)=(1+x+x^2+\dots+x^{n-1})-(x+x^2+x^3+\dots+x^n),$$ and we can simplify the right-hand side to $1-x^n$ since almost all of the terms cancel out.

Two aspects: One is to see that $\frac{1}{i}$ can be represented as integral. Note, that \begin{align*} \int_{0}^{1}x^{i-1}dx=\left.\frac{1}{i}x^{i}\right|_0^1=\frac{1}{i}\qquad\qquad i\geq 1 \end{align*} The other is to use the formula for the finite geometric series \begin{align*} \sum_{i=0}^{n}x^{i}=\frac{1-x^{n+1}}{1-x}\qquad\qquad n\geq 0 \end{align*}

These two aspects together with the linearity of the integral operator give \begin{align*} H_n&=\sum_{i=1}^{n}\frac{1}{i}=\sum_{i=1}^{n}\int_{0}^{1}x^{i-1}dx =\int_{0}^{1}\sum_{i=1}^{n}x^{i-1}dx\\ &=\int_{0}^{1}\sum_{i=0}^{n-1}x^{i}dx=\int_{0}^{1}\frac{1-x^n}{1-x}dx\qquad n\geq 1 \end{align*}