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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?

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    $\begingroup$ 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. $\endgroup$ – Suzet Jun 28 '18 at 14:04
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    $\begingroup$ A sequence of irrational numbers does not necessarily converge to an irrational number. $\endgroup$ – D. Brogan Jun 28 '18 at 14:05
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    $\begingroup$ $\sqrt{2}/n$ is irrational for each $n$- can I conclude that its limit, $0$, must be irrational too? $\endgroup$ – user281392 Jun 28 '18 at 14:06
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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.

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