The estimation of Harmonic numbers

I am following "Concrete Mathematics"'s treatment of Harmonic Numbers (pg. 262 onwards). They show that $\lim_{n\to\infty}\left(H_{n}-\ln n\right)=1-\frac{1}{2}\left(\zeta\left(2\right)-1\right)-\frac{1}{3}\left(\zeta\left(3\right)-1\right)-\dots$ using some beautiful arguments, and then claim the right hand side converges (and is the famous Euler's constant). However, I don't understand how they can claim it converges without further justification, which I can't find.

Indeed, they show just before that $\ln n<H_n<\ln n+1$; however, this also does not imply convergence, unless I'm missing something.

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In the second edition, it's on p. 278, and they say "In fact, $\zeta(r)-1$ is approximately $1/2^r$, so this infinite series converges rather rapidly and we can compute the decimal value $\gamma=0.5772156649\ldots$". Maybe they added that as a clarification in the second edition? – Hans Lundmark Jun 19 '11 at 17:24
No, now I see I have this as well... – Gadi A Jun 19 '11 at 17:29

1 Answer

I believe I have found an answer: To show that $1-\frac{1}{2}\left(\zeta\left(2\right)-1\right)-\frac{1}{3}\left(\zeta\left(3\right)-1\right)-\dots$ converges it is enough to show that $\sum_{k=2}^{\infty}\left(\zeta\left(k\right)-1\right)$ converges, but exchanging the order of summation (legal, since the series is positive) we have $\sum_{n=2}^{\infty}\left(\sum_{k=2}^{\infty}\frac{1}{n^{k}}\right)=\sum_{n=2}^{\infty}\frac{1}{n\left(n-1\right)}$ which obviously converges (the last step using geometric series formula).

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To connect with the hint in the text, here's another way of doing it: $$\zeta(r)-1=\frac{1}{2^r} + \sum_{n=3}^{\infty} \frac{1}{n^r} <\frac{1}{2^r} + \int_{2}^{\infty} \frac{dx}{x^r} = \frac{1}{2^r} \left( 1 + \frac{1}{r-1} \right) < \frac{1}{2^{r-1}}.$$ – Hans Lundmark Jun 19 '11 at 19:21