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I wanted to prove as much as I could about the rate of divergence of the harmonic series without resorting to textbooks; I did this by checking a little computationally and using that as motivation for the next bit. It wasn't very hard to prove the following

$$ \sum_{n=1}^N \frac{1}{n} = \log N + \gamma + O\Big(\frac{1}{N}\Big). $$

I then wanted an estimate for the implied constant above, and extensive computations showed that for all $N \in \mathbb{N}$,

$$ N\bigg( \sum_{n=1}^N \frac{1}{n} - \log N - \gamma\bigg) < \frac{1}{2} \qquad\qquad\qquad (1) $$

although this quantity does converge to $\frac{1}{2}$ from below (I plotted this sequence up to $N=10^{200}$ or so). This suggests that $C=\frac{1}{2}$ is the best implied constant we can get. Now, Example 2.1.10 here seems to imply

$$ N\bigg( \sum_{n=1}^N \frac{1}{n} - \log N - \gamma\bigg) > \frac{1}{2}, \qquad\qquad\qquad (2) $$

at least for all $N$ large enough.

Question: I am confused -- why is there an apparent contradiction in (1) and (2)?

(If you are not convinced by equation (1) I'd encourage you to try as many values of $N$ until you are convinced, or otherwise find me an integer value such that it does not hold.)

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    How did you prove (0) please?2011-05-06
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    I think the problem here is just that $10^{200}$ is peanuts compared to how far you'd have to look to see the harmonic series do *anything*. That is, the problem is with your claim (1).2011-05-06
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    The sum of the harmonic series to $10^200$ is roughly $430$, that means that if you have a double-logarithmic progression, it would only be around $6.5$ and so on. At four consecutive $\ln$ you reach to $~0.6$. I do not know how fast this sequence you describe converges, but if you can't *really* conclude anything from simply checking all the way to $10^200$, in fact I can easily come up with a series which is so slow you would think it converges by any current "ultrafinitistic method" (i.e. checking up to a high enough limit), but in fact it would diverge to infinity.2011-05-06
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    I believe (1) is correct. The sign in front of $1/12 x^2$ is incorrect in the linked document.2011-05-06
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    Fabian is right. The linked document has the correct minus sign in front of $1/12x^2$ in the proof; they just lost it when moving to the statement of the result.2011-05-06
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    @quanta: I showed that the sequence $a_N = \sum_{n=1}^N \frac{1}{n} - \log N$ is monotone and bounded above, hence convergent, and defined $\gamma := \lim a_N$. A little bit of thinking should then give you that $\sum_{n=1}^N \frac{1}{n} - \log N -\gamma = \int_{N}^\infty \frac{x-[x]}{x^2} dx < \int_{N}^\infty \frac{1}{x^2} dx = \frac{1}{N}$.2011-05-06
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    Ah, thanks Fabian and Mike, that puts my mind at ease.2011-05-06
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    @Fabian: Good catch. I am just inherently distrustful of reasoning by "checking lots of numbers", so I posted without checking the accuracy of the claim in the pdf.2011-05-06
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    @Asaf: I am not sure what your comment is intended to imply. I mean, what you say is absolutely correct: one can conclude nothing from checking up to $10^{200}$, and given any big number to check up to / "ultrafinitistic method", a series can be constructed that will fool it. However, it sounds like you think this is unreasonable, and that one *should* be able to conclude something by checking up to $10^{200}$. Perhaps I am misreading your tone (this being a textual medium and all).2011-05-06
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    @Zev: You surely misread my tone if you thought I meant that checking to $10^n$ for some finite $n$ (arbitrarily large) is enough. :-)2011-05-06
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    @Asaf: My sincere apologies :) I think I was thrown off by the "if" in "but if you can't *really* conclude...".2011-05-06

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Euler Maclaurin's formula shows that $H_N = \log N + \gamma + C_1/N + \ldots + C_k/N^k + O(1/N^{k+1})$ for constants $C_i$, and specifically, $H_N = \log N + \gamma + 1/(2N) - 1/(12N^2) + O(1/N^3)$, so $N\bigg( \sum_{n=1}^N \frac{1}{n} - \log N - \gamma\bigg)$ converges to $1/2$, and is less than $1/2$ for sufficiently large $N$.