Home > Analysis, Measure Theory > Sleek proof that the harmonic series diverges

## Sleek proof that the harmonic series diverges

Suppose that the harmonic series does converge. Then the sequence of integrable functions functions $f_n : \Bbb{R}_+ \to \Bbb{R}_+,\ f_n=\frac{1}{n} \chi_{[0,n]}$ is bounded above by an integrable function $g: \Bbb{R}_+ \to \Bbb{R}_+,\displaystyle \ g=\sum_{i=1}^\infty \frac{1}{i} \chi_{[i-1,i]}$.

Therefore, since $f_n \to 0$ almost everywhere, the Dominated Convergence theorem applies and we have

$0=\int_0^\infty \lim_{n \to \infty}f_n = \lim_{n \to \infty}\int_0^\infty f_n=1,$

and we have reached a contradiction. Therefore the harmonic series diverges.

In the following blog post there are two links in which over 40 ideas of proof are presented, all different from the one presented here.

Categories: Analysis, Measure Theory
1. April 24, 2012 at 8:27 am

Is there a similar argument for:

$\sum_{i=1}^\infty \frac{1}{n^2}$

2. April 24, 2012 at 12:23 pm

The proof works for proving the divercence of $\sum_{i=1}^\infty \frac{1}{n^\alpha}$ for $\alpha < 1$. I don't think that the proof can be adapted to prove the convergence of the harmonic series for $\alpha > 1$, because it assumes the convergence in order to reach a contradiction.

3. November 4, 2013 at 2:21 pm

But why wouldn’t the same proof work for $\sum \frac{1}{2^n}$? We assume it converges, then we get a contradiction.

• November 4, 2013 at 2:40 pm

I believe you want to say: $\sum \frac{1}{n^2}$.

If we define $f_n = \frac{1}{n^2} \chi_{[0,n]}$ then the integral of $f_n$ over $[0,n]$ is $1/n$ which converges to zero, so we don’t have a contradiction.

• November 4, 2013 at 6:34 pm

What I meant was defining $f_n=\frac{1}{2^n}\chi_{[0,2^n]}$, then the integral of $f_n$ is 1.

4. November 4, 2013 at 8:03 pm

Jj :

What I meant was defining $f_n=\frac{1}{2^n}\chi_{[0,2^n]}$, then the integral of $f_n$ is 1.

Then you have $g = \sup_n f_n = \sum_{n \geq 1} \frac{1}{2^n} \chi_{(2^{n-1},2^n]}$, and the integral of $g$ is $\sum_{n \geq 0} \displaystyle \frac{2^n -2^{n-1}}{2^n}=\sum_{n\geq 0} \frac{1}{2}\ (\star)$ which is divergent. This means that you cannot bound $f_n$ by an integrable function $g$ so Lebesgue dominated theorem does not apply.

Note that the difference between $\sum_{n \geq 1} \frac{1}{n}$ and $\sum_{n \geq 1}\frac{1}{2^n}$ is that the difference between consecutive terms is $1$ for the first one, and $2^n-2^{n-1}$ for the second one and that difference appears at the denominator in $(\star)$.

• November 4, 2013 at 8:12 pm

Thanks, I appreciate it.