I obtained the very strange formula above (in the question title) and I'm \*starting to understand* what does it mean. It's very similar to Poisson summation and indeed it is possible to get Poisson summation from that form  <br> 
* First of all, I point out that the above formula cannot be used for numerical calculations because of its divergence.
* If I understand it correctly, to sum a series from  $-\infty $ to $\infty $ means to broke it in two parts at some point $a$ and to treat each of them as an analytic continuation: if my understanding is not correct and/or you know anything about it or what does it mean, please let me know.

**Proof:** Let $f(x)$ be some an analytic function with a given Taylor series. By using the theorem on divergent summation derived by Ramanujan, we can associate to it the convergent integral $\int_x^{\infty}f (t) dt $ so we can write the summation in terms of Euler-Maclaurin formula as follows 
$$
\sum_{k=x}^{\infty}f (k)=- \sum_{n=0}^{\infty} \frac {(-1)^nf^{(n-1)}(x)B_n}{n!}.
$$
We know also that 
$$
(-1)^nB_n=2 \pi \int_{-\infty}^{\infty} \frac {(-\frac {1}{2}+it)  ^n}{(e^{\pi t}+e^{-\pi t})^2}dt, 
$$ 
so we get 
$$
\begin{split}
\sum_{k=x}^{\infty} f (k) &=-2 \pi \sum_{n=0}^{\infty}\frac{f^{(n-1)} (x)}{n!}\int_{-\infty}^{\infty} \frac {(-\frac {1}{2}+it)  ^n}{(e^{\pi t}+e^{-\pi t})^2}dt\\ 
& \text{ and by using Taylor series }\\
&=-2\pi\int_{-\infty}^{\infty} \frac {F(x-\frac {1}{2}+it)}{(e^{\pi t}+e^{-\pi t})^2}dt 
\end{split}$$

Let's use the consequences of Ramanujan's theorem another time
$$
\begin{split} 
\sum_{k=x}^{\infty} f (k) & = -2\pi\int_{-\infty}^{\infty} \frac {F(x-\frac {1}{2}+it)}{(e^{\pi t}+e^{-\pi t})^2}dt\\
& = 2\pi \int_{\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\int_{0}^{x-\frac {1}{2}+it}\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(u)^k}{k!}du}{(e^{\pi t}+e^{-\pi t})^2}dt\\
& = 2\pi \int_{-\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(x-\frac {1}{2}+it)^{k+1}}{k!(k+1)}du}{(e^{\pi t}+e^{-\pi t})^2}dt\\
& = \int_{0}^{\infty}f(u)du+\sum_{k=0}^{\infty}\frac{f^{(k)}(0)\zeta (-k,x)}{k!}
\end{split}
$$ 
Now let's assume $x=1$: then we can use Riemann's zeta functional equation to derived some transformation of the above expression:
$$
\begin{split} 
\sum_{k=1}^{\infty} f (k) & = \int_{0}^{\infty}f(t)dt+ \sum_{k=0}^{\infty}\frac{f^{(k)}(0)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \Gamma (1+k)\zeta (1+k) }{k!} \\ 
& =  \int_{0}^{\infty}f(t)dt+\sum_{n=1}^{\infty}\sum_{k=0}^{\infty} \frac { f^{(k)}(0)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) n^{-k-1}\Gamma (k+1)}{k!} \\ 
& = \int_{0}^{\infty}f(t)dt+ \sum_{n=1}^{\infty}\int_{0}^{\infty}\sum_{k=0}^{\infty}\frac{f^{(k)}(0) t^{k}(e^{-2 \pi i nt}+e^{2 \pi i nt})}{k!}dt\\
& =   \int_{0}^{\infty}f(t)dt+ \sum_{n=1}^{\infty}\int_{0}^{\infty}f(t)(e^{-2 \pi i nt}+e^{2 \pi i nt})dt\\
\end{split}
$$ 
This is enough to write down the formula as 
$$ \sum_{k=1}^{\infty} f (k)=\sum_{n=-\infty}^{\infty} \mathcal{L}  \{ f \} (2 \pi i n)$$

To contrast there is formula of Poisson summation, where $\hat {f}(x)=\int_{-\infty}^{\infty}f (t) e^{-2 \pi i xt}dt$ 


$$\sum_{n=-\infty}^{\infty} f (n)= \sum_{n=-\infty}^{\infty} \hat { f} (n)$$

**Edit:** Proof of Poisson summation from my theorem

Let $x $ be an integer number 

$\sum_{n=1-x}^{\infty} f (n)=\sum_{n=1}^{\infty} f (n-x)= \sum_{n=-\infty}^{\infty} \int_{0}^{\infty}f(t-x)e^{-2 \pi i nt}dt $

For integer $x $ we can write down $\lim\limits_{x \rightarrow \infty}\sum_{n=1-x}^{\infty} f (n)=\lim\limits_{x  \rightarrow \infty}\sum_{n=-\infty}^{\infty}\int_{0}^{\infty}f(t-x)e^{-2 \pi i nt}dt=\sum_{n=-\infty}^{\infty}\int_{-\infty}^{\infty}f(t)e^{-2 \pi i nt}dt$

Faze of exponent do not change because  $x $ is integer