I've noticed the exponential function $e^{-z}$ can be evaluated as a sum over $\log$ terms as follows where the evaluation limits $N$ and $f$ are both assumed to be positive integers.
$$e^{-z}=\underset{N,f\to\infty}{\text{lim}}\left(-\frac{1}{2} \sum\limits_{n=1}^N \frac{\mu(2 n-1)}{2 n-1} \sum\limits_{k=1}^{2 f\, (2 n-1)} (-1)^k\, \log\left(\frac{\pi^2\, k^2}{(2 n-1)^2\, z^2}+1\right)\right)\tag{1}$$
Formula (1) above was derived via the Mellin convolution
$$e^{-z}=-\int\limits_0^\infty \delta'(x-1)\, \text{Ei}(-z\, x)\, dx\tag{2}$$
where the convolution was evaluated using an analytic representation of $\delta'(x-1)$.
I believe formula (1) above is valid at least for $z>0$ if not for $\Re(z)>0$ which I illustrate following the question below.
Question: Can the $\log(z)$ function be evaluated as a sum over exponential terms?
Figure (1) below illustrates formula (1) for $e^{-z}$ evaluated at $N=50$ in orange and $N=100$ in green overlaid on the blue reference function where formula (1) is evaluated for $z>0$ and using the evaluation limit $f=4$ for both orange and green curves.
Figure (1): Illustration of formula (1) for $e^{-z}$ for $z>0$
Figures (2) and (3) below illustrate the real and imaginary parts of formula (1) for $e^{-z}$ evaluated at $N=50$ in orange and $N=100$ in green overlaid on the blue reference function where formula (1) is evaluated along the line $z=\frac{1}{2}+i\,t$ and using the evaluation limit $f=4$ for both orange and green curves in both figures.
Figure (2): Illustration of real part of formula (1) for $e^{-z}$ evaluated along the line $z=\frac{1}{2}+i\,t$
Figure (3): Illustration of imaginary part of formula (1) for $e^{-z}$ evaluated along the line $z=\frac{1}{2}+i\,t$
I've noticed that formula (1) above seems to converge better when the evaluation limit $N$ is selected such that the sum
$$\sum\limits_{n=1}^N \frac{\mu(2 n-1)}{2 n-1}\tag{3}$$
evaluates close to zero and note that
$$\sum\limits_{n=1}^\infty \frac{\mu(2 n-1)}{2 n-1}=\frac{1}{\lambda(1)}=0\tag{4}$$
where $\lambda(s)$ is the Dirichlet lambda function.
To clarify my question above, I'm not interested in a Taylor series, or a periodic Fourier series that converges over some small interval, but rather something more analogous to the following nested exponential series:
$$\frac{z}{z^2+1}=\underset{N,f\to\infty}{\text{lim}}\left(\frac{\pi}{2} \sum\limits_{n=1}^N \frac{\mu(2 n-1)}{2 n-1} \left(\frac{1}{2}+\sum\limits_{k=1}^{2 f\, (2 n-1)} (-1)^k\, e^{-\frac{\pi k z}{2 n-1}}\right)\right),\quad\Re(z)>0\tag{5}$$