From the definition of Euler's constant I knew that the following is true
$$\lim_{N \to \infty} \left( \sum_{n=1}^{N} \frac{1}{n}-\log N \right)=\gamma$$
I decided to check for other similar divergent series with Mathematica and got the following results:
$$\lim_{N \to \infty} \left( \sum_{n=1}^{N} \frac{1}{\sqrt{n}}-2\sqrt{N} \right)=\zeta \left( \frac{1}{2} \right)=-1.46035$$
$$\lim_{N \to \infty} \left( \sum_{n=1}^{N} \frac{1}{\sqrt[3]{n}}-\frac{3}{2} N^{2/3} \right)=\zeta \left( \frac{1}{3} \right)=-0.97336$$
And for the convergent series this is trivially true, since any integral of $a_n$ is $0$ in the limit:
$$\lim_{N \to \infty} \left( \sum_{n=1}^{N} \frac{1}{n^2}+\frac{1}{N} \right)=\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$$
I think this may be why the integral convergence test works.
So does this work for every divergent series? Can we just 'regularize' all of them in this way?
And by the way, why isn't $\zeta(1)=\gamma$?
Note that for each of the series you tried, $a_n=f(n)$ for a decreasing continuous function $f(x)$ on $[1,\infty)$. In general, there is no reason for an arbitrary sequence $a_n$ to have such regular behavior. But in this special case, the sum and integral are very closely related by the integral test, as you've observed.