Is there anything wrong with this proposed proof of the irrationality of Euler's constant?

Let $\{\lambda_n\}$ be the sequence given by $H_n - \ln n$. We claim that $\lambda_n$ is irrational for every integer $n>1$ and justify this by the following argument:

Assume that $\lambda_k$ is rational for some integer $k>1$ such that $H_k - \ln k = p/q$ where $p$ and $q$ are integers.

Rearranging the above we arrive at $H_k - p/q = \ln k$, which implies that $\ln k$ is rational since $H_k$ is rational. But we know that $\ln k$ is irrational for all integers $k>1$, hence we reach a contradiction.

Therefore, $\lambda_n$ is irrational for all integers $n>1$. Hence the limit as $n$ tends to infinity is irrational, and we are done.

You are implicitly claiming that the set of irrational numbers is closed in $\mathbb R$, which it is not. Indeed its closure is $\mathbb R$ (it is a dense subset), i.e. every real number is the limit of a sequence of irrational numbers.

• Thank you all for your comments, this is a really fascinating site ! And thanks Micah for the editions. – user264948 Aug 27 '15 at 21:51

You can have a rational limit of a sequence of all irrational numbers. Consider for example $\sqrt{2{\sqrt{2{\sqrt{2\ldots}}}}} = 2$

• Well in this case, the type of definition of the sequence is different, isn't it ? May you please come up with a sequence whose definition is similar to that of Euler's constant ? – user264948 Aug 27 '15 at 21:33
• It would perhaps be too hard to define "similar" in the definition of Euler's constant in a way that could give a counter-example and satisfy your want for similarity. But if you look at the abstraction of your statement, in general, for any given real number $x$, whether rational or irrational, there are uncountably many possible sequences of purely irrational numbers $x_n$ that converge to $x$. – user2566092 Aug 27 '15 at 21:36
• @Tatenda: $\displaystyle \sum_{n=1}^{N}\frac{1}{n^2} - \frac{\pi^2}{6}$ is irrational for every $N$, but the sequence converges to zero. – Bungo Aug 27 '15 at 21:40
• @CarstenS: The simplest way to see it is probably by noticing that the logarithm of left hand side is just $\log(2)\cdot (\sum_{n\geq 1} 2^{-n})$ – tomasz Aug 28 '15 at 1:49
• @tomasz Or note that, if $x = \sqrt{2\sqrt{2\sqrt{2...}}}$, then $x = \sqrt{2x}$ – user2103480 Aug 28 '15 at 11:56

For example, $$\lambda_n := e - \sum_{k=0}^n\frac{1}{k!}$$ are all irrational numbers, but their limit is zero.

• $\lambda_n := x/n$ also works, for any irrational $x$. – chi Aug 28 '15 at 6:24