25
$\begingroup$

Is this a valid proof that the harmonic series diverges?

  1. Assume the series converges to a value, S:

$$S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...$$

  1. Split the series into two, with alternating even and odd denominators. Since the original series converges, the component series will converge.

$$S_{EVEN}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...$$ $$S_{ODD}=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...$$ $$S=S_{EVEN}+S_{ODD}$$

  1. Show that $S_{EVEN}=\frac{1}{2}S$

$$\frac{1}{2}S=\frac{1}{2}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...)=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...=S_{EVEN}$$

  1. Show $S_{ODD}>S_{EVEN}$ because each odd term is greater than its corresponding even term: $$1>\frac{1}{2}\qquad \frac{1}{3}>\frac{1}{4}\qquad \frac{1}{5}>\frac{1}{6}\qquad ...$$

  2. Show $S_{ODD}=S_{EVEN}$ $$S_{ODD}=S-S_{EVEN}=S-\frac{1}{2}S=\frac{1}{2}S=S_{EVEN}$$

  3. The contradiction implies that the original assumption of convergence is false:

$$S_{ODD}>S_{EVEN}$$ $$S_{ODD}=S_{EVEN}$$ $$\therefore S\ne 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...$$

$\endgroup$
  • 3
    $\begingroup$ I think this is valid but I’ll let someone with more expertise weigh in on it. $\endgroup$ – Clayton Mar 20 at 14:46
  • 15
    $\begingroup$ It is valid. I would like to see a comment that all the terms are positive, so if it converges it converges absolutely, which justifies splitting the series in two. If the series did not converge absolutely splitting the series would not be valid. $\endgroup$ – Ross Millikan Mar 20 at 15:03
  • 6
    $\begingroup$ A surprising and elegant proof (+1) ! $\endgroup$ – Peter Mar 20 at 15:07
  • 2
    $\begingroup$ This is beautiful. $\endgroup$ – zugzug Mar 20 at 19:29
  • 2
    $\begingroup$ +1, Compare web.williams.edu/Mathematics/lg5/harmonic.pdf $\endgroup$ – Dr. Wolfgang Hintze Mar 21 at 16:37
12
$\begingroup$

This is almost valid. We need to justify the second step, as mentioned by Ross Millikan, as it is not always valid to split a series into their even and odd terms.

Take, as a simple example, the alternating harmonic series, where you would get it equalling $\infty-\infty$ which is indeterminate, but it does not make sense for the convergence of a series to be indeterminate.

This can be justified by seeing your series is absolutely convergent, assuming it converges.

If one must be pedantic, the same issue occurs showing $S_\mathrm{ODD}>S_\mathrm{EVEN}$, but this can be more easily justified by the fact we are comparing terms in the order that they are summed. If they were not compared in this order and their respective series converged conditionally, this may not be true.

Aside from all that it looks good. If I may provide an alternative proof of similar approach, it would've sufficed to have shown that

$$S=1+\frac12+\frac13+\frac14+\dots>\frac12+\frac12+\frac14+\frac14+\dots=S$$

$\endgroup$
3
$\begingroup$

Rearrangement requires knowledge of absolute convergence. When I wrote up my proof and it was published 23 years ago, that was the comment added. Other than that, your proof is absolutely identical to mine. Here is the reference:

Michael W. Ecker, Divergence Of The Harmonic Series By Rearrangement, The College Mathematics Journal, May 1997, Vol. 28, No. 3, p. 209-210.

Several years later, Bernard August and Thomas Osler cited this and generalize this method in the May 2002 issue of The College Mathematics Journal, p. 233-234. If memory serves, they applied this to other p-series, but I don't have the issue in front of me.

$\endgroup$
  • 1
    $\begingroup$ Thanks. I've looked up your paper and found it on ResearchGate. $\endgroup$ – Adam Hrankowski Mar 25 at 20:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.