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It is known that $\displaystyle\sum_1^{\infty} \frac{1}{n^{1.000001}}$ converges while $\displaystyle\sum_{n\text{ is a prime number}}\frac{1}{n}$ diverges. Though we can logically prove these results, I found it difficult to explain their intuition to my students. How could it be done?

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    $\begingroup$ I think the proofs are lovely (iirc). And I'll be surprised if there's anything better. $\endgroup$
    – Sue VanHattum
    Nov 10, 2023 at 23:08

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Look at a simpler example first: $(1.000000000001)^n$ compared to $0.9999999999^n$. Do they accept that the first sequence tends to $\infty$ and the second to $0$ even though it would take quite a while to see either such behavior happening? Not all interesting mathematical properties in a sequence have to occur among the first 100, 1000, or even million terms.

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For me, the intuition just comes from the integral test (which is itself intuitive since a series is just a Riemann sum of rectangles with unit width).

  • The $n$th prime is asymptotically $n \ln n$ (see here for intuition on that), and the integral of $\dfrac{1}{x \ln x}$ is $\ln \ln x,$ whose end behavior is divergence as $x \to \infty.$

  • But as soon as that exponent in the denominator exceeds $1,$ the integral is a reciprocal power function, whose end behavior is convergence as $x \to \infty.$

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    $\begingroup$ OK this works, but the OP is asking for an intuitive explanation. The fact that $p_n$ is like $n\ln n$ relies on the Prime Number Theorem, which is not intuitive! Honestly, I downvoted the original question because the two series are so unrelated...seems fake to be trying to explain both to students. Had the second series been the standard harmonic series then the question would not seem so absurd. $\endgroup$
    – user52817
    Nov 11, 2023 at 2:59
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    $\begingroup$ @RichardHardy no, assuming you start at $n=2$ to avoid the blowup in the first term, it diverges by the integral test because the integral yields $\ln \ln n.$ $\endgroup$ Nov 11, 2023 at 20:34
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    $\begingroup$ @Richard Hardy: The $p$-series and the (harmonic) logarithmic $p$-series (for real numbers $p,$ each converges for each $p>0$ and each diverges for each $p \leq 1)$ are the first two levels of an infinite sequence of iterated logarithmic convergence/divergence scales that I believe was first introduced by Abel (1820s; not sure off-hand the precise year), although I find it surprising that this isn’t something that Euler had previously found. Maybe Abel was the first to provide rigorous proofs? (continued) $\endgroup$ Nov 12, 2023 at 21:54
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    $\begingroup$ Two common and relatively straightforward ways to show their convergence/divergence is the integral test and the Cauchy condensation test. This scale is mentioned in several of the books I listed in this MSE answer (e.g. pp. 52-53 of Bonar/Khoury, p. 40 of Hyslop (1954 5th edition), pp. 122-123 of Knopp’s 1954 book and pp. 62-63 of Knopp’s 1956 book, etc.), and sometimes in more theoretical/advanced introductory calculus texts, and sometimes in advanced calculus texts, and sometimes in real analysis texts, etc. (continued) $\endgroup$ Nov 12, 2023 at 21:54
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    $\begingroup$ Useful (and freely available, at least at the present time) is Series involving iterated logarithms by Ash (2009). See also the MSE question/answer Convergence of series involving in iterated logarithms $\sum \frac{1}{n(\log n)^{\alpha_1}\cdots (\log^k(n))^{\alpha_k} }$. An excellent reference for those teaching calculus and undergraduate real analysis is Bonar/Khoury’s 2006 book (see this MSE answer for a nice summary of their book). $\endgroup$ Nov 12, 2023 at 22:01
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Intuitively, to me, it means that if you take the positive number line, put a blue dot at every prime, and a red dot on all the the numbers of the form $n^{1.000000000001}$, then eventually, very far out on the number line, there will be more blue dots than red dots.

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  • $\begingroup$ I agree with your explanation. To me, "there will be more blue dots than red dots" eventually is mind-blowing. $\endgroup$
    – Zuriel
    Nov 13, 2023 at 15:33
  • $\begingroup$ @Zuriel I remember having my mind blown when I came to a similar conclusion about $1.01^n$ many, many years ago. It happens to all of us at some point. $\endgroup$
    – Arthur
    Nov 13, 2023 at 16:50
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Consider the fact that $\sum_{n=1}^\infty n^x$ converges if $x<0$, diverges if $x>0$. Clearly the transition from just a little bit negative to just a little bit positive makes a big change to the sum. This may not help your students see why $x=0$ diverges, but it should at least help them not to be surprised that it behaves differently.

Afterword. I forgot to mention that $\sum_{n=1}^\infty n^x = \zeta(-x)$, and the Riemann zeta function has a pole at $x=1$. That may put it in conext for students who want to explore.

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    $\begingroup$ I don't follow the first sentence. Actually, the series converges for $x<-1$ and diverges for $x\ge-1$. $\endgroup$ Nov 13, 2023 at 8:41

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