2
$\begingroup$

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 '14 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. $\endgroup$ – Thomas Andrews May 22 '14 at 0:21
  • $\begingroup$ @ThomasAndrews I am sure you have a typo there. $\endgroup$ – chubakueno May 22 '14 at 0:23
  • $\begingroup$ Fixed. @chubakueno $\endgroup$ – Thomas Andrews May 22 '14 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 '14 at 3:58
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 '14 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. $\endgroup$ – Thomas Andrews May 22 '14 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 '14 at 4:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.