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.

Source: http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/44742#44742

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.

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  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. Jj
    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.

      • Jj
        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).

    • Jj
      November 4, 2013 at 8:12 pm

      Thanks, I appreciate it.

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