Why is this proof of the irrationality of $\gamma$ wrong?

Question

By definition,

$$\gamma=\lim_{n\to \infty}\left(\sum_{k=1}^n \left(\frac{1}{k}\right)-ln(n)\right)$$

It is clear that $\sum_{k=1}^n \left(\frac{1}{k}\right)$ will always be rational, as it is a sum of rational numbers.

It can be showed that $\ln(n)$ is always irrational.

Thus, as the difference of a rational number and an irrational number is always irrational, it would follow that $\gamma$ could never be rational; therefore, $\gamma$ is irrational.

$0$ is the limit when $n$ grows to infinity of $\frac{1}{n+\sqrt{2}}-\frac{1}{n}$, which is a sequence of irrational numbers. — Aphelli, Dec 05 '19 at 19:38
With this logic how about saying $\displaystyle e = \sum_{n=0}^\infty \frac{1}{n!}$ is rational? — Feel_Free_To_Edit_This_Answer, Dec 05 '19 at 19:41

score 1 · Answer 1 · answered Dec 05 '19 at 19:44

1

Yes, $\gamma$ is the limit of a sequence of irrational numbers. So is $1$, since$$1=\lim_{N\to\infty}\sum_{n=0}^N\frac{e^{-1}}{n!}.$$However, $1$ is rational.

answered Dec 05 '19 at 19:44

José Carlos Santos

427,504

A simpler example could be $0 = \lim\limits_{n\rightarrow \infty} \frac{1}{\sqrt{p_n}}$ where $p_n$ is the $n$-th prime number. – WhatsUp Dec 05 '19 at 19:46
Sure. Or even $0=\lim_{n\to\infty}\frac1{\sqrt{n^2+1}}$. – José Carlos Santos Dec 05 '19 at 20:15
Or the definition of square root of 2 as the sup of a sequence of rational numbers whose squares are less than 2 – lux Dec 05 '19 at 20:15
Thanks for your great answers! – Juan Moreno Dec 06 '19 at 09:48

score 0 · Answer 2 · answered Dec 05 '19 at 19:44

The limit of a sequence of irrational numbers could be any real number. In particular it could be rational. This is true just as it is true that the limit of a sequence of rational numbers can be irrational. This is true for most useful series expansions for $\pi$, $e$, etc.

score 0 · Answer 3 · answered Dec 05 '19 at 19:48

0

Is $ln(n)$ always irrational on the domain $[1, \infty)$? What about $ln(1)$? The first approximation $$\gamma = \frac{1}{1} - ln(1) = 1$$ is certainly not irrational.

answered Dec 05 '19 at 19:48

Ty Jensen

1,516

Why is this proof of the irrationality of $\gamma$ wrong?

3 Answers3