I obtained the very strange formula above and at begining I was just wanted know how to interpretate it. But now when I know what is this with help of @Carlo Beenakker, I am leaving it as a proof. BTW I found some article about non formal take of that theorem (I hope my take on this could be colled formal)

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 $$ \begin{split} \sum_{k=x}^{\infty} f (k) & = -2\pi\int_{-\infty}^{\infty} \frac {F(x-\frac {1}{2}+it)}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = 2\pi \int_{\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\int_{0}^{x-\frac {1}{2}+it}\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(u)^k}{k!}du}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = 2\pi \int_{-\infty}^{\infty} \frac {\int_{0}^{\infty}f(u)du-\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(x-\frac {1}{2}+it)^{k+1}}{k!(k+1)}du}{(e^{\pi t}+e^{-\pi t})^2}dt\\ & = \int_{0}^{\infty}f(u)du+\sum_{k=0}^{\infty}\frac{f^{(k)}(0)\zeta (-k,x)}{k!}\\ &=\sum_{k=-1}^{\infty}\frac{f^{(k)}(0)\zeta (-k,x)}{k!} \end{split} $$ The form above proves that Taylor series i truely defined at least as $f (n)=\sum_{k=a}^{\infty}\frac{f^{(k)}(0)n^k}{k!} \wedge a <0$ because for divergent summtion theorem $\sum_{n=x}^{\infty}f (n)=\sum_{k=-1}^{\infty}\frac{f^{(k)}(0)\zeta (x,-k)}{k!} $

Let's assume $x=1$: then we can use Riemann's zeta functional equation to derived some transformation of the above expression: $$ \begin{split} \sum_{k=1}^{\infty} f (k) & = \int_{0}^{\infty}f(t)dt+ \sum_{k=a}^{\infty}\frac{f^{(k)}(0)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) \Gamma (1+k)\zeta (1+k) }{k!} \\ & = \int_{0}^{\infty}f(t)dt+\sum_{n=1}^{\infty}\left (\sum_{k=0}^{\infty}+\sum_{a}^{-1} \frac { f^{(k)}(0)(2 \pi)^{-k-1} (i^{-k-1}+(-i)^{-k-1}) n^{-k-1}\Gamma (k+1)}{k!} \right) \\ & = \int_{0}^{\infty}f(t)dt+ \sum_{n=1}^{\infty}\int_{0}^{\infty}\sum_{k=0}^{\infty}\frac{f^{(k)}(0) t^{k}(e^{-2 \pi i nt}+e^{2 \pi i nt})}{k!}dt+\Omega (0) \\ & = \int_{0}^{\infty}f(t)dt+ \sum_{n=1}^{\infty}\int_{0}^{\infty}f(t)(e^{-2 \pi i nt}+e^{2 \pi i nt})dt+\Omega (0). \end{split} $$ This is enough to write down the formula as $$ \sum_{k=1}^{\infty} f (k)=\sum_{n=-\infty}^{\infty} \mathcal{L} \{ f \} (2 \pi i n)+\Omega (0)$$

From Poisson formula we know that $$\sum_{n=-\infty}^{\infty} f (n)= \sum_{n=-\infty}^{\infty} \hat { f} (n)$$, which is sufficent to identyfify $\Omega (0) $. In such way we finaly get

$$ \displaystyle \frac {f (0)}{2}+ \sum_{k=1}^{\infty} f (k)=\sum_{n=-\infty}^{\infty} \mathcal{L} \{ f \} (2 \pi i n).$$