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Today I entered the following expression in maple: $$a_i = H_{10^i} - ln(10^i) - \gamma$$ Here $H_j$ equals $\sum_{k=1}^{j} 1/k$ and $\gamma$ is the Euler-Mascheroni constant.

When I computed $a_n$ for $i = 0$ to $10$ I obtained the following results:

  • $i = 0$;    4.227843350984671393934879099175975689578406640600764011942327651151323 * $10^{-1}$
  • $i=1$;   4.9167496072675423629464709201487329610707429399557393414873118115813 * $10^{-2}$
  • $i=2$;   4.991666749996032162622676207122311664609813510982102304110919767206 * $10^{-3}$
  • $i=3;$   4.999166666749999960317501984051226762153678825611388678758121701133 * $10^{-4}$
  • $i=4$;   4.99991666666674999999960317460734126976551226762154503821179264423 * $10^{-5}$
  • $i=5$;   4.9999916666666667499999999960317460321626984126226551226762154523 * $10^{-6}$
  • $i=6$;   4.999999166666666666749999999999960317460317501984126984051226523 * $10^{-7}$
  • $i=7$;   4.99999991666666666666674999999999999960317460317460734126984123 * $10^{-8}$
  • $i=8$;   4.9999999916666666666666667499999999999999960317460317460321623 * $10^{-9}$
  • $i=9$;   4.999999999166666666666666666749999999999999999960317460317423 * $10^{-10}$
  • $i=10$; 4.99999999991666666666666666666674999999999999999999960317423 * $10^{-11}$

So we see that the periodic strips of ...99999..., of ...66666... and ...99999... an many other periods increase for even larger $i$. The question is now: Is there any rule behind it that the remainder term $a_i$ behaves that way?

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Curious - have you examined the base $b$ expansion of $H_{b^i}- \ln(b^i) - \gamma$? I rather doubt $10$ is special here, and perhaps the more general perspective will shed some light on the matter. –  Ramsey Mar 12 '12 at 14:26
@Ramsey: yes one would see analog phenomena in any base, as you expect; see my answer and the comment to it (eventually/soon to be merged/expanded/clarifed). –  quid Mar 12 '12 at 17:22
(I replaced \begin{itemize} etc., which doesn't work in this environment, with <ul> etc., which does.) –  Joseph O'Rourke Mar 12 '12 at 20:45
Surely a general base can be considered, I just found the basis 10 the most canonical one to consider this interesting behaviour. –  tobias Mar 19 '12 at 18:33
The following blog post on Euler's constant on Lipton-Regan's blog featuring this paper of Jeff Lagarias might be of some relevance. –  Gil Kalai Dec 29 '13 at 19:26

2 Answers 2

up vote 25 down vote accepted

Yes, there is a rule. There are results that are finer than merely $H_k - \ln k - \gamma$ tends to $0$ and explain this pattern.

More specifically, let us consider some more terms of the asymtotic expansion of $H_k$ . One has for example $$H_k = \ln k + \gamma + \frac{1}{2k} - \frac{1}{12k^2} + O(k^{-4}) $$ and this is even true with a small implied constant, or more precisely this is true with $O(k^{-4})$ replaced by $x_k\frac{1}{120}k^{-4}$ with $0 \le x_k \le 1$. Thus the error to be expected when doing the calculation in the question is $$\frac{1}{2k} - \frac{1}{12k^2}$$ up to something still (much) smaller.

This is precisely what one sees; if one chooses for $k$ a power of $10$ one sees a nice pattern (the $10$ being special due to the fact that one has the decimal representation; if one chooses a different base for the representation, powers of that base become special); it is the beginning of the decimal representation of $$\frac{1}{2} 10^{-j} - \frac{1}{12} 10^{-2j} ;$$ how long it is really just this can also be known from the estimate of the error mentioned above.

One can continue on this, as it is known that $$H_k = \ln k + \gamma + \frac{1}{2k} - \sum_{i=1}^{n-1} \frac{B_{2i}}{2i k^{2i}} + O(k^{-2n})$$ and more precisely the $O(n^{-2k})$ can be replaced by $x_{k,n}( -\frac{B_{2n}}{2n}) k^{-2n} $ with $0\le x_{k,n} \le 1$ where the $B$'s are the Bernoulli numbers; some care is needed if one would want to try to see more complex patterns as the Bernoulli numbers while small at first then grow very fast, so that then the implied constant is large and the $k$ needs to be sufficiently large (relative to the $n$) to see the pattern for all the terms.

Besides the approximation I mentiond above there are various other approximations known. Also, questions like this are closely linked, essentially equivalent, to questions on the Digamma function .

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How does this explain the OP's observation? –  Igor Rivin Mar 12 '12 at 13:40
@Igor Rivin: the observed errors are all very sligthly smaller 5 times 10^(-j-1) matching closely the 1/2(10^j). And the next term in the exxpansion explains why the are all slightly smaller, and by about how much. In other words, the expansion I mention explains that the error will 5 times 10^(-j-1) minus something still alot smaller. And you can also understand who much and one can also 'see' the 1/12 . And if one goes further one will 'see' the other terms very nicely at powers of 10 in base ten, or any other power in the matching digital expansion. –  quid Mar 12 '12 at 14:31
To explain the above comments: the original version was shorter. And, perhaps I should not have stressed so much (unfortunaletly I still do this) the two term version as one indeed sees already more terms. –  quid Mar 12 '12 at 19:26
@Igor Rivin: perhaps you see it easier if you subtract 1 from the value, say at i=6, and then divide by the first few bernoulli numbers, say $\small (a_6−1)/{1 \over∗6} $or $\small (a_6−1) / {1 \over∗42} $ or even by two bernoulli-numbers $\small (a_6−1)/{1 \over 42}/{1 \over 30 } $ –  Gottfried Helms Mar 12 '12 at 22:15

Adding an example to @quid's answer:

Using Pari/GP the harmonic numbers minus Euler-gamma can be obtained using the psi-function. If we subtract 1 from the $\small a_i $-values the composition of the result by i-digits long blocks of decimal expansion of the bernoulli-numbers becomes then immediately visible. With $\small i \gt 20 $ or so it becomes even more impressive:

fmt(200,60)   \\ user function: internal prec 200, display prec 60 digits
psi(10^i)-1 - i*log(10)
 %681 = -1.00000050000008333333333332500000000000396825396824980158730
(psi(10^i)-1 - i*log(10))*6
 %683 = -6.00000300000049999999999995000000000002380952380949880952381
(psi(10^i)-1 - i*log(10))*42
 %684 = -42.0000210000034999999999996500000000001666666666664916666667
(psi(10^i)-1 - i*log(10))*42*30
 %685 = -1260.00063000010499999999998950000000000499999999999475000000


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