A better approximation of $H_n $

Question

I'm convinced that

$$H_n \approx\log(n+\gamma) +\gamma$$ is a better approximation of the $n$-th harmonic number than the classical $$H_n \approx \log(n) +\gamma$$ Specially for small values of $n$. I leave some values and the error:

Just to make things clearer, I calculate the value between two numbers as follows. Say $n$ is the optimal and $a$ is the apporximation, then $E = \frac{n-a}{n}$. $L_1$ stands for my approximation and $L_2$ for the classical one, and the errors $E_2$ and $E_1$ correspond to each of those (I mixed up the numbers).

It is clear that this gives an over estimate but tends to the real value for larger $n$.

So, is there a way to prove that the approximation is better?

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NOTE: I tried using the \begin{tabular} environment but nothing seemed to work. Any links on table making in this site?

Answer 1

Actually, you do better still with $ H_n \approx \gamma + \log \left( n + \frac{1}{2} \right),$ with $$ H_n = \gamma + \log \left( n + \frac{1}{2} \right) + O \left( \frac{1}{n^2} \right). $$ As you can see from the other answers, this minimizes the error among approximations of type $H_n \approx \gamma + \log \left( n + c \right)$ with constant $c,$ by erasing the $\frac{1}{n} $ error term.

A fuller version of the asymptotic above is just

$$ H_n = \gamma + \log \left( n + \frac{1}{2} \right) + \frac{1}{24 \left( n + \frac{1}{2} \right)^2} - \frac{7}{960 \left( n + \frac{1}{2} \right)^4} + \frac{31}{8064 \left( n + \frac{1}{2} \right)^6} - \frac{127}{30720 \left( n + \frac{1}{2} \right)^8} + O \left( \frac{1}{n^{10}} \right). $$

Answer 2

The asymptotic expansion of the Harmonic numbers $H_n$ is given by

$$\log n+\gamma+\frac{1}{2n}+\mathcal{O}\left(\frac{1}{n^2}\right).$$

The Maclaurin series expansion of the natural logarithm tells us $\log(1+x)=x+\mathcal{O}(x^2)$, and we can use this in your formula by writing $\log(n+\epsilon)=\log n+\log(1+\epsilon/n)$ and expanding:

$$\log(n+\gamma)+\gamma=\log n+\gamma\;\;\;+\frac{\gamma}{n}+\mathcal{O}\left(\frac{1}{n^2}\right).$$

Your approximation is asymptotically better than the generic one because the $\gamma=0.577\dots$ in its expansion is closer to the true coefficient $\frac{1}{2}$ than the illicit coefficient $0$ in the generic formula given by the usual $H_n\sim \log n +\gamma+0/n$. This also explains why it is asymptotically an over estimation.

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As marty said in his answer, the expansion comes from the Euler-Maclaurin formula:

$$\sum_{n=a}^b f(n)=\int_a^b f(x)dx+\frac{f(a)+f(b)}{2}+\sum_{k=1}^\infty \frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(b)-f^{(2k-1)}(a)\right).$$

Here we let $a=1,b=n$ (rewrite the index to a different letter) and $f(x)=1/x$.

Answer 3

The true approximation, from the Euler-Maclaurin formula, is $$H_n = \ln n + \gamma + 1/2n + O(1/n^2).$$

Your expansion is $\ln (n+\gamma) + \gamma = \ln \ n + ln(1+\gamma/n)+\gamma = \ln \ n + \gamma + \gamma/n + O(1/n^2) $.

Since $\gamma = .577...$, your error is about $.077/n$, which is better by .077/.5 ~ .154 ~ 1/6 which explains the table.

I see another answer has just been posted. If it differs much from mine, I will be surprised.