Is there another proof for Euler–Mascheroni Constant?

Question

Problem

Prove that the sequence $$x_n=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n,~~~(n=1,2,\cdots)$$is convergent.

One Proof

This proof is based on the following inequality

$$\frac{1}{n+1}<\ln \left(1+\frac{1}{n}\right)<\frac{1}{n}$$

where $n=1,2,\cdots$, which will be used repeatedly.

On one hand, we obtain that $$\ln 2-\ln 1<1,~~\ln 3-\ln 2<\frac{1}{2},~~\ln 4-\ln 3<\frac{1}{3},~~\cdots,~~\ln (n+1)-\ln n<\frac{1}{n}.$$ Adding up all of these，we have that $\ln(n+1)<1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}.$ Hence,$$x_{n+1}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)>\frac{1}{n+1}>0.$$ This shows that $x_n$ is bounded below.

On the other hand,$$x_n-x_{n+1}=-\frac{1}{n+1}+\ln(n+1)-\ln n=\ln \left(1+\frac{1}{n}\right)-\frac{1}{n+1}>0.$$ This shows that $x_n$ is decreasing. Combining the two aspects, according to Monotone Bounded Theorem, we can assert that $\lim\limits_{n \to \infty}x_n$ exists.

Let $\gamma$ (so-called Euler–Mascheroni Constant) denote the limit, i.e. $$\gamma=\lim_{n \to \infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\ln n\right),$$which equals $0.577216 \cdots$. We may also express that as $$1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\gamma+\ln n+\varepsilon_n,$$where $\varepsilon_n$ represents an infinitesimal related to $n$ under the process $n \to \infty$.

Answer 1

Pictorial proof.

In the picture, take $n=11$. The red graph is $1/x$, so the area under the red graph from $1$ to $n$ is $$ \int_1^n\frac{dx}{x} = \ln n. $$ The area of the white rectangles under the graph is $$ \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} $$ The difference is shown in green, $$ \text{area}(\text{green}_n) = \ln(n) - \left(\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right) $$ Take the green triangles, translate them to the left, so they are in the strip between $x=0$ and $x=1$. They are disjoint. So their areas add to at most the area of the whole strip $(0,1) \times (0,1)$; that is $1$...
$$ \text{area}(\text{green}_n) < 1 $$ Now as $n$ increases, we add more green regions, so this value increases with $n$, and it bounded above by $1$, so it converges and has limit $\le 1$:
$$ 0 < \lim_{n\to\infty}\left[\ln(n) - \left(\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right)\right] \le 1 $$ The value you want is obtained by subtracting all of this from $1$:
$$ 0 \le \lim_{n\to\infty}\left[ \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right)- \ln(n)\right] < 1 $$

If we like, we can think of this as the sum of the gray areas.

These little gray and green regions are approximately triangles... if they were exactly triangles, then our result would be exactly $1/2$. As it is, $\gamma$ is approximately $1/2$.

Answer 2

You can prove it by showing it is strictly decreasing and bounded from below, by the Maclaurin-Cauchy theorem, which states:

If $f\colon[0;+\infty)\longrightarrow \mathbb{R}$ is positive, continuos, strictly decreasing and $\lim_{x \to \infty}f(x)=0$, then $$\gamma_{f}=\lim_{n \to \infty} \bigg(\sum_{i=1}^{n}f(i) - \int_1^n f(x)\, \mathrm{d}x\bigg)$$ exists, where $\gamma_{f}$ is an Euler constant defined by $f(x)$.

A proof of this theorem:

It follows from hypothesis that

$$f\, \text{continuos }\implies \forall \: n \geq 1 \: \exists \: I_{n}=\int_1^n f(x)\, \mathrm{d}x$$

and that

$$f \text{ strictly decreasing} \implies \inf_{k; k+1}f(x) = f(k+1) \vee \sup_{k; k+1} f(x) = f(k)$$

which implies $$f(k+1)\leq f(x) \leq f(k) \forall x \mid k \leq x \leq k+1.$$

Then integrating from $k$ to $k+1$: $$\int_{k}^{k+1} f(k+1)\,\mathrm{d}x \leq \int_{k}^{k+1} f(x)\,\mathrm{d}x \leq \int_{k}^{k+1} f(k)\,\mathrm{d}x$$

but as only the middle term of the inequality is not a constant, it follows that $$f(k+1) \leq \int_{k}^{k+1} f(x)\, \mathrm{d}x \leq f(k).$$

We now sum from $k=1$ to $k=n-1$ so that the the integral becomess $I_{n}$ from : $$\sum_{i=2}^{n}\, f(i) \leq I_{n} \leq \sum_{1}^{n-1} f(i)$$

Then we try to find $a_{n}$ by rewriting: $$\sum_{i=1}^{n}f(i)-f(1) \leq I_{n} \leq \sum_{i=1}^{n}f(i) - f(n)$$

Inverting the signs and adding $\sum_{i=1}^{n}f(i)$, we get $$-\sum_{i=1}^{n}f(i)+f(1) \geq -I_{n} \geq -\sum_{i=1}^{n}f(i) + f(n)$$

$$f(1)\geq a_{n} \geq f(n).$$

Since $f(x)$ is positive by hypothesis, this shows that $a_{n}\geq 0\:$and is thus bounded below.

To show that $a_{n}$ is decreasing, we consider the difference of the term $n+1$ from its preceding: $$a_{n+1}-a_{n}=\sum_{i=1}^{n+1}f(i)-\int_{1}^{n+1}f(x)\,\mathrm{d}x - \bigg(\sum_{i=1}^{n}f(i)-\int_{1}^{n}f(x)\,\mathrm{d}x\bigg)=$$

$$f(n+1)-\int_{n}^{n+1}f(x)\,\mathrm{d}x$$

Then, we observe that by a previous equation $$f(n+1)-\int_{n}^{n+1}f(x)\,\mathrm{d}x\leq 0.$$

Hence

$$a_{n+1}-a_{n}\leq 0 \rightarrow a_{n+1}\leq a_{n}.$$

As the above shows that $a_{n}$ is decreasing and we've shown it is bounded from below, so by your Monotone Bounded Theorem $a_{n}$ is convergent. $\Box$

Then if you take $f(x)=1/x$ we have

$$\gamma_{\frac{1}{x}}=\lim_{n \to \infty}\bigg(\sum_{i=1}^{n}\frac{1}{i}-\int{1}^{n}\frac{1}{x}\,\mathrm{d}x\bigg)=$$

$$=\lim_{n \to\infty}\bigg(\sum_{i=1}^{n}H_{i}-\ln{n}+\ln{1}\bigg)=$$

$$=\lim_{n \to \infty}\bigg(\sum_{i=1}^{n}H_{i}-\ln{n}\bigg),$$

which is Euler's constant.

(Sorry if my answer is not clearly formatted, this is my first contribution on this site and i tried to adapt a typeset I did in $\LaTeX$)

Answer 3

It follows from Riemann-Stieltjes integral that

$$ \sum_{k=1}^n\frac1k=1+\int_1^n{\mathrm d\lfloor x\rfloor\over x}=\log n+1-\int_1^n{\{x\}\over x^2}\mathrm dx $$

Consequently, all we need is to show that the integral

$$ \int_1^n{\{x\}\over x^2}\mathrm dx $$

converges as $n\to\infty$, and this trivially follows from the fact that $\{x\}\le1$.

Answer 4

Let the sum of the reciprocals be $H(n)$. Then: $$(H(n+1)-H(n))-(\ln(n+1)-\ln(n))=\frac{1}{n+1}-\ln\left(\frac{n+1}{n}\right)$$This means that the sequence is decreasing. Now all you need to prove is that $\dfrac{1}{\lfloor x \rfloor}>\dfrac{1}{x}$, which could be easily done using induction.