In another question of mine, I found a regulator $f$ such that $f(s,0) = 1$ and

\begin{align}
\lim_{\varepsilon \rightarrow 0^+}\lim_{m \rightarrow \infty} \sum_{n=1}^m n^s f(s, n \varepsilon) &= \zeta(-s)
\end{align}

for all $s \neq -1$, namely
\begin{align}
f(s,x) &= \mathrm{e}^{-x}\left(1 - \frac{x}{s+1}\right)
\end{align}

Hence we can also regularize the harmonic series as follows:
\begin{align}
\gamma
&= \frac{1}{2} \lim_{\varepsilon \rightarrow 0^+} (\zeta(1+\varepsilon) + \zeta(1-\varepsilon)) \\
&= \frac{1}{2} \lim_{\varepsilon \rightarrow 0^+} \left(\lim_{m \rightarrow \infty} \sum_{n=1}^m n^{-1-\varepsilon} f(-1-\varepsilon, n \varepsilon) + \lim_{m \rightarrow \infty} \sum_{n=1}^m n^{-1+\varepsilon} f(-1+\varepsilon, n \varepsilon) \right) \\
&= \frac{1}{2} \lim_{\varepsilon \rightarrow 0^+} \lim_{m \rightarrow \infty} \left(\sum_{n=1}^m n^{-1-\varepsilon} f(-1-\varepsilon, n \varepsilon) + \sum_{n=1}^m n^{-1+\varepsilon} f(-1+\varepsilon, n \varepsilon) \right) \\
&\overset{\star}{=} \frac{1}{2} \lim_{m \rightarrow \infty} \lim_{\varepsilon \rightarrow 0^+} \left(\sum_{n=1}^m n^{-1-\varepsilon} f(-1-\varepsilon, n \varepsilon) + \sum_{n=1}^m n^{-1+\varepsilon} f(-1+\varepsilon, n \varepsilon) \right) \\
&= \frac{1}{2} \lim_{m \rightarrow \infty} \left(\sum_{n=1}^m n^{-1} f(-1, 0) + \sum_{n=1}^m n^{-1} f(-1, 0) \right) \\
&= \frac{1}{2} \lim_{m \rightarrow \infty} \left(\sum_{n=1}^m n^{-1} + \sum_{n=1}^m n^{-1} \right) \\
&= \lim_{m \rightarrow \infty} \sum_{n=1}^m n^{-1} \\
&= \sum_{n=1}^\infty n^{-1} \\
\end{align}

where the star indicates the non-rigorous step of exchanging limits. Here is the Mathematica code and its plots:

```
f[s_, x_] := Exp[-x] (1 + a x)
g[s_, t_] :=
Evaluate@Simplify[
f[s, t] /. Solve[Integrate[x^s f[s, x], {x, 0, Infinity}] == 0, a],
Assumptions -> Re[s] > -1]
g[s, t]
Table[{s,
Plot[{Zeta[-s],
Sum[n^s g[s, n \[Epsilon]], {n, 1, 1000}]}, {\[Epsilon], 0, 1},
Evaluated -> True]}, {s, -4, 4, 1/2}] // TableForm // Quiet
Plot[{EulerGamma,
Sum[(n^(-1 + \[Epsilon]) g[-1 + \[Epsilon], n \[Epsilon]] +
n^(-1 - \[Epsilon]) g[-1 - \[Epsilon], n \[Epsilon]])/
2, {n, 1, 1000}]}, {\[Epsilon], 0, 1}, Evaluated -> True]
```