Convergence of sequence (Euler - Mascheroni constant)

Let $\mathcal{H}_n$ denote the $n$-th harmonic number. Prove that the sequence:

$$\gamma_n = \mathcal{H}_n - \ln n$$

converges.

Solution

In order to prove that the sequence converges we have to show two things. Firstly that is monotonic and secondly that is bounded.

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.

The limit of this sequence is a very well known constant, namely Euler - Mascheroni constant. More information on this constant can be found here