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.

Answer 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.

Answer 2

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.

Answer 3

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.