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The proof involving partial sums up to the nth term, where n is some power of $2$, completely makes sense. But just looking at the series itself, it seems very strange that it's divergent.

For large values of $n$, $a_n$ would start being extremely small and having an indistinguishable effect on the overall sum. All the sixth sense I've gained from working with limits makes it seem really strange that this would be considered divergent.

Surely there is a number (not even that difficult to find) such that we don't have enough computational power to calculate it's difference with the next terms (seeing as we'd be calculating differences based on hundreds of decimal places).

If you've any intuition on this I'd very much love to hear it!

Edit: I'm not asking for the proof of why it's divergent, I'm asking for peoples' personal ways of thinking and making sense of this intuitively. The post suggested to have been duplicated presents formal proofs; that's not what I'm looking for :)

James Ronald
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  • Intitutive reason: https://www.youtube.com/watch?v=aKl7Gwh297c – Sujit Bhattacharyya Mar 30 '19 at 13:15
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    There are a very large number of very small values. The question is which very effect dominates when adding these up (in a sense you are multiplying very large by very small), and it seems intuitively plausible that in some cases very large number can be more important than very small value but not in other cases – Henry Mar 30 '19 at 13:16
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    In Analysis, our professor told us for intuitively understanding those things, we have to think in the way "that series are approximately integrals : $\sum \approx \int$". So we see $ \int_1^\infty \frac{\mathrm{d}x}{x} = \ln(x)\vert_1^\infty = \infty$. On the other hand, we know that the series $\sum_k \frac{1}{k^2}$ converges. Considering an integral, we have $\int_1^\infty \frac{\mathrm{d}x}{x^2} = - \frac{1}{x}\vert_1^\infty = 1 < \infty$. For me, this idea is good because I understand integrals better than series. – Jan Mar 30 '19 at 13:22
  • Consider the following intuition,

    Going into hyperreals, note that there exists $\epsilon$ infinitesimal that is greater than any $1/n$.

    Then harmonic series 'sum to' $\omega \cdot \epsilon =1$, where $\omega := 1/\epsilon$ , 'approximately' is the number of terms.

    For other $p$-series, where $p\gt 1$, they 'sum to' $\omega \cdot 1/\epsilon ^p = \epsilon^{1-p}$ which is still infinitesimal. Hence making a difference.

     –  Mar 30 '19 at 13:28
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    I think the connection to the logarithm brought up by @Jan is important. For high values of n, the partial sums are approximately log(n) + γ. And log(n) is proportional to the number of digits of n, ±1. So your question is like, "when n is a very large number, adding 1 almost certainly has no effect on the number of digits of n. Doesn't that mean that there's a maximum number of digits?" And the answer is no, because you can increase the number of digits by multiplying by 10--just like how the proof you mentioned does *2. – user54038 Mar 30 '19 at 13:33

2 Answers2

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From Real Infinite Series by Bonar:

We know that, for $x>-1$, $$x \geq \ln(1+x)$$ Now $$\sum_1^n \frac{1}{k}\geq \sum_1^n \ln\left(1+\frac{1}{k}\right)=\ln(n+1) \longrightarrow \infty$$ as $n \to \infty$ and hence the divergence of the harmonic series follows

We can interpret this argument in a much more strikingly visual way as follows:

Consider the following graph of the function $g(x) = \sin(\pi e^x)$, shown below, We consider $g$ as a function only of positive reals, We know that this function is defined for arbitrarily large $x$. We also know that $\sin x$ is zero at integer multiples of $\pi$, so that $g$ has zeros whenever $e^x$ is integer-valued, which happens of course for $x$ of the form $\log n$. The distance between consecutive zeros is of the form $\log(k + 1) — \log k$, which by the argument above is a lower bound to $1/k$. This is the motivation for the choice of the function $g$—the oscillations make visible the segments between zeros, and the lengths of these segments estimate the terms of the harmonic series. If the harmonic series were to converge to some number $N$, then the length sum of all the segments between zeros of $g$, since they are smaller, would also be bounded above by $N$. Then $g$ could have no further zeros right of the vertical line $x = N$, but we know this does not happen. Again we emphasize that this contains no mathematical content not present in the argument above, only a new way to make it tangible.

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Added: Also the author of the above mentioned book gives $11$ proofs of " $\sum \frac{1}{n}$ is divergent". So refer this book for more details!

Chinnapparaj R
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Variation of the proof cited by the OP: let be $S_n$ the $n$-th partial sum: $$S_{2n} - S_n = \sum_{k=n+1}^{2n}\frac{1}k\ge(2n - n)\frac{1}{2n} = \frac{1}2.$$ (lower bound: number of terms times smallest term)

David C. Ullrich
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Martín-Blas Pérez Pinilla
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