# A better approximation of $H_n$

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?

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

• LaTeX text formatting commands usually don't work, but you could put your numbers in a \begin{array} or \begin{matrix}. It might be somewhat slow. – Dylan Moreland Feb 3 '12 at 3:14
• I might be misunderstanding, but if you're trying to compute "relative error" of the two approximations, shouldn't you be doing $\left|\dfrac{(\text{approximation})-(\text{true value})}{\text{true value}}\right|$? From that, you can try to look at the asymptotic behavior of the two errors as $n\to\infty$... – J. M. is a poor mathematician Feb 3 '12 at 3:25
• That's what I'm doing. I chose the largest first to get a positive number. – Pedro Tamaroff Feb 3 '12 at 12:01
• – Martin Sleziak Aug 4 '12 at 8:59
• @MartinSleziak Good to know. – Pedro Tamaroff Aug 4 '12 at 14:37

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

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

• $n+1/2$ is certainly easier to deal with than $n+\gamma$... :) – J. M. is a poor mathematician Feb 3 '12 at 3:47
• @WillJagy Where do I find theory on such expansion? – Pedro Tamaroff Feb 3 '12 at 20:02
• @Peter, the standard asymptotic (in powers of $\frac{1}{n}$) comes from the relationship with the digamma function, see en.wikipedia.org/wiki/… and the first formula in the section en.wikipedia.org/wiki/… I did the version with adding $(1/2)$ myself, but it is surely equivalent to the other. There is a good reason that adding $(1/2)$ will erase the first error term in any similar problem, but the explanation is intricate. – Will Jagy Feb 3 '12 at 21:27
• Ok, thanks for the info. – Pedro Tamaroff Feb 4 '12 at 16:46
• $H_n\approx log\left(2n+1\right)+\gamma-log(2)$ is better because $log(2)-\gamma < \gamma$. See math.stackexchange.com/a/1602945/134791 for a series linking $H_n$ and $log\left(2n+1\right)$ – Jaume Oliver Lafont Jan 27 '16 at 21:41

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.

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 $f(x)=1/x$.

• Maybe you can remove al the colours now. Asaf will be thankfull. – Pedro Tamaroff May 15 '12 at 16:28
• As you wish. ${}$ – anon May 15 '12 at 18:40

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.