If you are looking to collect other (truly) exotic solutions to this admittedly simple sum I can offer one using harmonic sums and Mellin transforms. This might well qualify as a winner where eclecticism is concerned.
Introduce $$T(x) = \sum_{n\ge 1}\frac{2n-1}{2^{xn}}
= \frac{1}{\sqrt{2}^x} \sum_{n\ge 1}\frac{2n-1}{2^{xn-x\times1/2}}
= \frac{1}{\sqrt{2}^x} \frac{1}{x} \sum_{n\ge 1}\frac{x(2n-1)}{\sqrt{2}^{x(2n-1)}}.$$
We will evaluate $T(x)$ by inverting the Mellin transform of
$$S(x) = \sum_{n\ge 1}\frac{x(2n-1)}{\sqrt{2}^{x(2n-1)}}$$ and then put $x=1$ to get the value of the sum.
Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = 1, \quad \mu_k = 2k-1 \quad \text{and} \quad
g(x) = x\sqrt{2}^{-x}.$$
We need the Mellin transform $g^*(s)$ of $g(x)$ which is
$$\int_0^\infty x\sqrt{2}^{-x} x^{s-1} dx =
\int_0^\infty \sqrt{2}^{-x} x^{(s+1)-1} dx
= \frac{1}{(1/2\log 2)^{s+1}}\Gamma(s+1).$$
Furthermore,
$$\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s}
= \sum_{k\ge 1} \frac{1}{(2k-1)^s} = (1-2^{-s})\zeta(s).$$
Therefore the Mellin transform $Q(s)$ of $S(x)$ is given by
$$Q(s) = \frac{1}{(1/2 \log 2)^{s+1}} (1-2^{-s}) \Gamma(s+1) \zeta(s).$$
The appropriate Mellin inversion integral is then given by
$$\frac{1}{2\pi i}\int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds.$$
The pole at $s=1$ from the zeta function term has residue
$$\mathrm{Res}(Q(s)/x^s; s=1) =
\frac{1}{x} \frac{1}{(1/2 \log 2)^2} (1-2^{-1})\Gamma(1)
= \frac{2}{x(\log 2)^2}.$$
The remaining residues of the poles at the negative integers are given by
$$- \sum_{q\ge 1} (1/2 \log 2)^{q-1} (1-2^q) \frac{(-1)^{q-1}}{(q-1)!}
\frac{B_{q+1}}{q+1} x^q.$$
The trivial zeros of the zeta function term at the negative even integers cancel the poles of the gamma function term at those values. We may still let $q$ run through all negative integers because the corresponding Bernoulli numbers are zero.
Re-write this as
$$- x \sum_{q\ge 1} (1/2 \log 2)^{q-1} (1-2^q) (-1)^{q-1}
\frac{B_{q+1}}{(q+1)!} q x^{q-1}.$$
Now the exponential generating function of the Bernoulli numbers is given by
$$\frac{t}{e^t-1} = \sum_{m\ge 0} B_m \frac{t^m}{m!}.$$
Therefore
$$-1 + \frac{t}{e^t-1} =
t \sum_{m\ge 1} B_m \frac{t^{m-1}}{m!}$$
and
$$\left(-\frac{1}{t} + \frac{1}{e^t-1}\right)'
= \sum_{m\ge 2} \frac{B_m}{m!} (m-1) t^{m-2}
= \sum_{m\ge 1} \frac{B_{m+1}}{(m+1)!} m t^{m-1}.$$
This gives
$$\frac{1}{t^2} - \frac{e^t}{(e^t-1)^2} =
\sum_{m\ge 1} \frac{B_{m+1}}{(m+1)!} m t^{m-1}.$$
We now substitute the appropriate values of $t$ from the sum of the residues into this formula, getting two terms, the first of which gives $t=-1/2\log 2 \times x$, producing
$$-x \left(\frac{1}{x^2 (1/2\log 2)^2}
- \frac{1/\sqrt{2}^x}{(1/\sqrt{2}^x-1)^2}\right).$$
The second term has $t=-\log 2 \times x$ and gives
$$-2x \left(\frac{1}{x^2 (\log 2)^2}
- \frac{1/2^x}{(1/2^x-1)^2}\right).$$
Collecting the contributions from all residues we finally obtain
$$ \frac{2}{x(\log 2)^2}
- \frac{1}{x (1/2\log 2)^2}
+ x\frac{1/\sqrt{2}^x}{(1/\sqrt{2}^x-1)^2}
+ \frac{2}{x (\log 2)^2}
- 2x\frac{1/2^x}{(1/2^x-1)^2}.$$
Carrying out the cancellation we have that
$$S(x) = x\frac{1/\sqrt{2}^x}{(1/\sqrt{2}^x-1)^2}
- 2x\frac{1/2^x}{(1/2^x-1)^2}.$$
This implies that
$$T(x) = \frac{1/2^x}{(1/\sqrt{2}^x-1)^2}
- \frac{2}{\sqrt{2}^x} \frac{1/2^x}{(1/2^x-1)^2}.$$
In particular we have for $x=1$ that
$$T(1) = \frac{1/2}{(1/\sqrt{2}-1)^2}
- \sqrt{2}\frac{1/2}{(1/2-1)^2}
= \frac{1/2 ((1/\sqrt{2}+1)^2)}{(1/2-1)^2}
- 2\sqrt{2}
\\ = 2 ((1/\sqrt{2}+1)^2) -2\sqrt{2}
= 2 (3/2+2/\sqrt{2}) -2\sqrt{2} = 3.$$
