# Irrationality of the Euler–Mascheroni constant

The definition of the Euler–Mascheroni constant is the limit of $$H_n - \log(n)$$ as n approaches infinity. So, why is it so hard to prove the irrationality of this constant? $H_n$ is defined only for integers and for any integer $n > 1$ , $\log(n)$ is irrational.

On the other hand, $H_n$ is always a rational number. Subtracting an irrational number from a rational one doesn't make the constant an irrational number?

• Yes, but why exactly would this argument holds after taking the limit on $n$ ? Consider for instance the sequence $\frac{1}{n}+\frac{\pi}{n}$. It is a sum of a sequence of rational numbers, and of a sequence of irrational numbers. Both sequence go to $0$, as well as their sum, and $0$ is a rational number. – Suzet Jun 28 '18 at 14:04
• A sequence of irrational numbers does not necessarily converge to an irrational number. – D. Brogan Jun 28 '18 at 14:05
• $\sqrt{2}/n$ is irrational for each $n$- can I conclude that its limit, $0$, must be irrational too? – user281392 Jun 28 '18 at 14:06

## 1 Answer

Consider the sequence $(x_n)_{n\in\mathbb N}$ defined by$$x_n=\log(2)-\sum_{k=1}^n\frac{(-1)^{k+1}}k.$$Each $x_n$ is an irrational number minus a rational number. However, $\lim_{n\to\infty}x_n=0$ and $0$ is rational.