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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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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. –  Asaf Karagila May 6 '11 at 20:54
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I believe (1) is correct. The sign in front of $1/12 x^2$ is incorrect in the linked document. –  Fabian May 6 '11 at 20:56
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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. –  Mike Spivey May 6 '11 at 21:00
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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. :-) –  Asaf Karagila May 6 '11 at 21:50
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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...". –  Zev Chonoles May 6 '11 at 21:55

1 Answer 1

up vote 11 down vote accepted

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 $N$.

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