3
$\begingroup$

On StackExchange, I read that the harmonic series up to $\frac{1}{n}$ is approximately $\ln(n) + \gamma$, where $\gamma$ is the Euler-Mascheroni constant, which is close to $0.5772$. When I researched the Euler-Mascheroni constant, I only found it defined in terms of the difference between the harmonic series and $\ln(n)$.

Why is the series able to be approximated in this way, and what is the Euler-Mascheroni constant?

$\endgroup$
5
  • 2
    $\begingroup$ Draw rectangles representing the sum (i.e. having height $1/n$ on the interval $[n, n + 1]$), and the area representing the integral. In the limit, these are almost the same, and the error is $\gamma$. $\endgroup$
    – user61527
    May 22, 2014 at 0:16
  • 5
    $\begingroup$ It's pretty easy to show, using $$\log n = \int_1^n\frac{dx}{x} = \sum_{i=1}^{n-1}\int_i^{i+1}\frac{dx}{x}$$ that $$\left(\sum_{i=1}^n\frac{1}{i}\right)-\log n$$ converges as $n\to\infty$. You can also easily prove that the limit is between $1/2$ and $1$. This limit, or some variant, is often the definition of the Euler-Mascheroni constant. $\endgroup$ May 22, 2014 at 0:21
  • $\begingroup$ @ThomasAndrews I am sure you have a typo there. $\endgroup$
    – chubakueno
    May 22, 2014 at 0:23
  • $\begingroup$ Fixed. @chubakueno $\endgroup$ May 22, 2014 at 0:24
  • $\begingroup$ I like showing $\gamma$ is between one increasing sequence and one decreasing sequence that are also forced to come together; see math.stackexchange.com/questions/306371/… $\endgroup$
    – Will Jagy
    May 22, 2014 at 3:58

2 Answers 2

7
$\begingroup$

$$\log n = \int_1^n \frac{dx}{x} = \sum_{i=1}^{n-1}\int_{i}^{i+1}\frac{dx}{x}$$

So:

$$\left(\sum_{i=1}^n \frac 1 i\right)-\log n = \left(\sum_{i=1}^{n-1}\int_i^{i+1}\left(\frac 1i-\frac1x\right)dx\right) + \frac{1}{n}$$

Now, for $x\in[i,i+1]$, $0\leq\frac{1}i-\frac1 x\leq \frac{1}{i(i+1)}.$

So these terms are positive and $\sum_{i=1}^\infty \frac{1}{i(i+1)} = 1$. So as $n\to\infty$, this means:

$$\left(\sum_{i=1}^n \frac 1 i\right)-\log n$$ converges to a value less than $1$.

It's actually pretty easy to show, since $f(x)=1/x$ is concave, that:

$$\int_i^{i+1}\left(\frac 1i-\frac1x\right)dx>\frac{1}{2i(i+1)}$$

This means that the limit is between $1/2$ and $1$.

This is often the definition of the Euler-Mascheroni constant.

$\endgroup$
2
  • $\begingroup$ Don't you need to show the series is increasing? Or have you shown that and I'm missing something. $\endgroup$
    – Alex Zorn
    May 22, 2014 at 0:49
  • $\begingroup$ If $0<a_i<b_i$ and $\sum b_i$ converges, then $\sum a_i$ converges. Basically, $\sum_{i=1}^n a_i$ is increasing as $n\to\infty$ and is bounded above, therefore converges. $\endgroup$ May 22, 2014 at 1:14
1
$\begingroup$

As noted in the comments, the "computational" answer is that, comparing areas, we find

$$\int_1^{n+1}\frac{dx}{x}<1+{1\over 2}+\cdots +{1\over n}<1+\int_1^n\frac{dx}{x}$$ Evaluating the integrals gives $$\ln(n+1)<1+{1\over 2}+\cdots +{1\over n}<1+\ln n$$ so we have the very rough approximation that, denoting $$H_n=1+{1\over 2}+\cdots +{1\over n}$$ the sum is around $H_n\approx \ln n$.


A more insightful perspective is to compare the nature of the functions $H_n$ (often called the "harmonic sum") and $\ln x$. The former is discrete, and the latter continuous. In "discrete calculus" you often see $$\Delta f(n)=f(n+1)-f(n)$$ which is the discrete analogue of the derivative. Note that the "derivative" of the harmonic sum is $$\Delta H_n=H_{n+1}-H_n=\left(1+\cdots +{1\over n+1}\right)-\left(1+\cdots +{1\over n}\right)={1\over n+1}$$

Similarly, the derivative of the logarithm is $$\textrm{D}\ln x={1\over x}$$

(The analogy can be drawn further, depending on how much discrete maths you know) There is a sense then, in which the two functions are "companions" of each other - they live in different universes, but within those universes they share many of the same properties. This is a recurring theme in much of mathematics (you may be interested in researching the word "isomorphic", although it does not completely apply here).

$\endgroup$
1
  • $\begingroup$ If you average $\ln (n + 1)$ and $\ln n + 1$ you get $\ln n + 0.5$ as an estimate for $H_n$ $\endgroup$
    – vonbrand
    May 22, 2014 at 4:36

You must log in to answer this question.