We can define the sum of the Harmonic series by
$$
\sum_{n=1}^\infty \frac{1}{n} = \frac{\eta'(1)}{\log(2)} = \gamma - \frac{\log(2)}{2}.
$$
A series $\sum_{n=1}^\infty b_n$ is Abel summable to $g(1)$ if (i) $g(x) = \sum_{n=1}^\infty b_n x^n$ converges for $|x| < 1$ (at least) and (ii) there exists the limit $g(1) = \lim_{x \to 1^-} g(x)$.
A series $\sum_{n=1}^{\infty} a_n$ belongs to the Ramanujan class $[R]$ if $f(x) = \sum_{n=1}^\infty a_n x^n$ and $f(x) - R f(x^2) = g(x)$, or rather, (i) $a_{2k-1} = b_{2k-1}$, $a_{2k} = b_{2k} + R a_k$ (so that $a_{2k}$ is a polynomial in $R$) and (ii) $\sum_{n=1}^\infty b_n$ is Abel summable.
For $R \neq 1$, $f(1) = \sum_{n=1}^\infty a_n$ is elementary Ramanujan summable to
$$
\frac{g(1)}{1 - R}.
$$
This summation is a linear function defined in $[R]$ (which is a linear space). It agrees with the analytic continuation of the Dirichlet series $F(s) = \sum_{n=1}^\infty a_n n^{-s}$ at $s=0$.
In fact, for $G(s) = \sum_{n=1}^\infty b_n n^{-s}$, the recursive relations between the coefficients $a_n$ and $b_n$ imply that
$(1 - R 2^{-s}) F(s) = G(s)$. Thus,
$$
F(s) = \frac{G(s)}{1 - R 2^{-s}}
$$
and
$$
F(0) = \frac{G(0)}{1 - R}.
$$
For $g(1) = G(0)$, $f(1) = F(0)$.
A summation to be defined in $[1]$ must be a linear function. If $g(1) = G(0) = 0$, the sum of $f(1) = F(0)$ is given by l'Hopital's rule,
$$
\sum_{n=1}^\infty a_n = \frac{G'(0)}{\log(2)}.
$$
For example, $f(x) = x$, $g(x) = x - x^2$, $F(s) = 1$, $G(s) = 1 - 2^{-s}$, $G'(0) = \log(2)$, $g(1) = G(0) = 0$, $f(1) = F(0) = 1$.
As the derivative $G'(0)$ is linear, the definition is natural in the entire linear space $[1]$.
The Harmonic series corresponds to $f(x) = - \log(1 - x) = \sum_{n=1}^\infty n^{-1} x^n$, $g(x) = \sum_{n=1}^\infty (-1)^{n-1} n^{-1} x^n$, $F(s) = \zeta(s+1) = \sum_{n=1}^\infty n^{-s - 1}$, $G(s) = \eta(s + 1) = \sum_{n=1}^\infty (-1)^{n-1} n^{-s - 1}$.
Thus, $R = 1$ and
$$
f(1) = F(0) = \frac{\eta'(1)}{\log(2)} = \gamma - \frac{\log(2)}{2}.
$$
In general,
$$
\sum_{n=1}^\infty a_n = \frac{G'(0)}{\log(2)} + G(0)\sum_{n=1}^\infty \text{pow2}(n),
$$
where $\text{pow2}(n)$ is $1$ if $n$ is a power of $2$, $0$ otherwise, and the sum of $\sum_{n=1}^\infty \text{pow2}(n)$ is set by definition. If $\sum_{n=1}^\infty \text{pow2}(n) = 1/2$, then $\sum_{n=1}^\infty 1/n = \gamma$.
This definition is natural, because
$$
P(s) = \sum_{n=1}^\infty \text{pow2}(n) n^{-s} = \frac{1}{1 - 2^{-s}}
$$
and
$$
\frac{P(s) + P(-s)}{2} = \frac{1}{2}.
$$