Suppose $\gamma$ is the Euler–Mascheroni constant .
Let $\gamma_n:=H_n-\ln(n)$.
Then $\gamma=\lim_{n \to \infty} \gamma_n.$
I have this summation $$\sum_{n=1}^{\infty}(\gamma_n-\gamma)$$
I would like to know if this summation converges or not.
If I calculate the limit $\lim_{n \to \infty}(\gamma_n-\gamma)$ it is clear that the result is zero. but it seems to me it diverges, so how to find the result if it converges and how to prove it.
EDIT
Now we are sure it diverges by using the harmonic formula shown in Iqcd answer.
But I think if we speed up $\gamma_n$ to go quickly to $\gamma$ it will converges by changing $n$ to $n^2$ so we have $\sum_{n=1}^{\infty}(\gamma_{(n^2)}-\gamma)$.
And it seems we have new function here diverges at $x=1$ $$f(x)=\sum_{n=1}^{\infty}(\gamma_{(n^x)}-\gamma)$$
at $x \to +\infty$ , $f(x)=1-\gamma$
We have (see Wikipedia)
$$\tag{$*$}\label{eq1}H_n=\log(n)+\gamma+\frac1{2n}+O(n^{-2})$$
so
$$\sum_{n\leq x}\left\{H_n-\log(n)-\gamma\right\} =\sum_{n\leq x}\left\{\frac1{2n}+O(n^{-2})\right\},$$
and therefore the series diverges.
ADDED (after the edit): let us define the function
$$f(x;N):=\sum_{n\leq N}\left\{H_{n^x}-\log(n^x)-\gamma\right\},$$
so that $f(x;N)\to f(x)$ when $N\to\infty$. Again by \eqref{eq1} we see that
$$f(x;N)=\sum_{n\leq N}\left\{\log(n^x)+\gamma+\frac1{2n^x}-\log(n^x)-\gamma+O(n^{-2x})\right\} =\sum_{n\leq N}\left\{\frac1{2n^x}+O(n^{-2x})\right\}.$$
Since $\sum n^{-p}$ converges iff $p>1$, we see that $f(x)$ converges iff $x>1$.