Euler-Macheroni constant

[Grigorios Kostakos]

Prove that the sequence \[\gamma_{n}=\displaystyle1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}-\log{n}\,, \quad n\in\mathbb{N}\,,\] converges.

[Papapetros Vaggelis]

For all \(\displaystyle{n\in\mathbb{N}}\) , \(\displaystyle{\gamma_{n}=\sum_{k=1}^n \frac{1}{k}-\log n}\) , so

\(\displaystyle{\begin{aligned} \gamma_{n+1}-\gamma_{n}&=\sum_{k=1}^{n+1}\frac{1}{k}-\log \left(n+1\right)-\sum_{k=1}^n \frac{1}{k}+\log n\\&=\sum_{k=1}^n \frac{1}{k}+\frac{1}{n+1}-\left[\log \left(n+1\right)-\log n\right]-\sum_{k=1}^n \frac{1}{n}\\&=\frac{1}{n+1}-\log \left(\frac{n+1}{n}\right)\end{aligned}}\)

Let \(\displaystyle{f:\left(0,+\infty\right)\to \mathbb{R}\,\,,f(x)=\log x}\).

The function \(\displaystyle{x\mapsto f^\prime(x)=\frac{1}{x}\,\,,x>0}\) is strictly decreasing at \(\displaystyle{\left(0,+\infty\right)}\) .

If \(\displaystyle{n\in\mathbb{N}}\), by applying the mean value theorem for \(\displaystyle{f}\), at the interval \(\displaystyle{\left[n,n+1\right]}\), we have that

there is \(\displaystyle{t\in\left(n,n+1\right)}\) such that

\(\displaystyle{f^\prime(t)=\frac{1}{t}=\frac{\log \left(n+1\right)-\log n}{\left(n+1\right)-n}=\log \left(\frac{n+1}{n}\right)}\) , so

\(\displaystyle{\frac{1}{n+1}<t<\frac{1}{n}\Rightarrow \frac{1}{n+1}<\log \left(\frac{n+1}{n}\right)<\frac{1}{n}}\)

Therefore, \(\displaystyle{\gamma_{n+1}-\gamma_{n}=\frac{1}{n+1}-\log \left(\frac{n+1}{n}\right)<0\ \forall n\in\mathbb{N}}\) , consequently,

the sequence \(\displaystyle{\left(\gamma_{n}\right)_{n\in\mathbb{N}}}\) is strictly decreasing.

Also, for all \(\displaystyle{n\in\mathbb{N}}\) ,

\(\displaystyle{\begin{aligned} \gamma_{n}=\sum_{k=1}^n \frac{1}{k}-\log n&\geq \sum_{k=1}^n \log \left(\frac{k+1}{k}\right)-\log n\\&=\log 2+\log \frac{3}{2}+\log \frac{4}{3}+...+\log \frac{n+1}{n}-\log n \\&=\log \left(2\cdot \frac{3}{2}\cdot \frac{4}{3}\cdot \cdot \cdot \frac{n+1}{n}\right)-\log n\\&=\log \left(n+1\right)-\log n\\&=\log \left(1+\frac{1}{n}\right)\\&>\log 1=0\end{aligned}}\) .

We proved that the sequence \(\displaystyle{\left(\gamma_{n}\right)_{n\in\mathbb{N}}}\) is bounded.

Thus, the sequence \(\displaystyle{\left(\gamma_{n}\right)_{n\in\mathbb{N}}}\) converges.

[Grigorios Kostakos]

Another solution:

For every \(n\in\mathbb{N}\) holds:
\begin{align*}
& \Bigl({1+\frac{1}{n}}\Bigr)^{n}<e<\Bigl({1+\frac{1}{n}}\Bigr)^{n+1}\quad\Rightarrow\quad\log\Bigl({\bigr({1+\tfrac{1}{n}}\bigr)^{n}}\Bigr)<\log{e}<\log\Bigl({\bigr({1+\tfrac{1}{n}}\bigr)^{n+1}}\Bigr)\quad\Rightarrow\\
& 0<n\log\tfrac{n+1}{n}<1<({n+1})\log\tfrac{n+1}{n}\quad\Rightarrow\quad\frac{1}{n\log\tfrac{n+1}{n}}>1>\frac{1}{({n+1})\log\tfrac{n+1}{n}}\quad\Rightarrow
\end{align*}\[\displaystyle\frac{1}{n+1}<\log\tfrac{n+1}{n}=\log({n+1})-\log{n}<\frac{1}{n}\quad(1)\,.\]
From \((1)\) we have\begin{align*}
& \log2-\log1<\frac{1}{1}=1\\
& \log3-\log2<\frac{1}{2}\\
& \dots\dots\dots\dots\dots\dots\dots\dots\dots\\
& \log n-\log({n-1})<\frac{1}{n-1}\\
& \log({n+1})-\log n<\frac{1}{n}
\end{align*}Adding the above equations, we have\begin{align*}
& \log({n+1})<\displaystyle1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}\quad\Rightarrow\\
& \log({n+1})-\log n<\displaystyle1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}-\log n=\gamma_{n}\,, \ n\in\mathbb{N}\,.
\end{align*}But for every \(n\in\mathbb{N}\), we have that \[\log({n+1})-\log n=\log\bigr({1+\tfrac{1}{n}}\bigr)>0\] and the sequence \(\left({\gamma_{n}}\right)_{n\in\mathbb{N}}\) is strictly decreasing because \begin{alignat*}{2}
\gamma_{n+1}-\gamma_{n} & =\displaystyle1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}+\frac{1}{n+1}-\log({n+1})-1-\frac{1}{2}-\frac{1}{3}-\ldots-\frac{1}{n}+\log n\\
& =\frac{1}{n+1}-\log\tfrac{n+1}{n}\stackrel{(1)}{<}0\,.
\end{alignat*} So, the sequence \(\left({\gamma_{n}}\right)_{n\in\mathbb{N}}\) converges to a real number \(\gamma\), named the Euler-Macheroni constant.

[Tolaso J Kos]

Splitting hairs around here, well, here is another approach that is not that much different than the previous two answers.

To prove that the sequence converges we have to show that it is bounded and monotonic.Here we go.

Monotony: It is easy to prove that the sequence is decreasing. Indeed:

$$\gamma_{n+1} - \gamma_n = \left ( \mathcal{H}_{n+1}- \ln (n+1) \right ) - \left ( \mathcal{H}_n - \ln n \right ) = \frac{1}{n}+ \ln \left ( 1- \frac{1}{n} \right )<0 $$

The inequality is valid since $\ln (1-x)$ is a concave function hence lies beneath the line $y=-x$ that is its tangent to its graph at $x_0=0$. Plugging $x=1/n$ yields the result.

Boundness: Using the fact that:

$$\mathcal{H}_n = \sum_{k=1}^{n}\frac{1}{k}> \int_{1}^{n}\frac{{\rm d}t}{t}= \ln (n+1)> \ln n$$

we have that $\gamma_n = \mathcal{H}_n -\ln n >0$. Hence $\gamma_n$ is bounded from below.


Therefore the sequence converges to the famous Euler - Mascheroni constant. More information can be found here