Fibonacci series captures Euler $e=2.718\dots$

It follows from the identity

$$F_{n+k} = \sum_{j=0}^k {k \choose j} F_{n-j}$$

which is obtained by applying the standard recurrence $k$ times to the left side, each time splitting up each term into two terms.

Indeed, this gives

$$\sum_{k=0}^{\infty} \frac{F_{n+k}}{k!} = \sum_{k=0}^{\infty} \frac{ \sum_{j=0}^k {k \choose j} F_{n-j}}{k!}= \sum_{k=0}^{\infty} \sum_{j=0}^{k} \frac{1}{j! (k-j)!} F_{n-j} = \sum_{j=0}^{\infty} \sum_{k-j=0}^{\infty} \frac{1}{(k-j)!} \frac{F_{n-j}}{j!} = \sum_{j=0}^{\infty} e \frac{F_{n-j}}{j!}$$

Mathematica tells me it's a consequence of the two series (distinguished by $\pm$): $$\sum_{k=0}^\infty\frac{F_{n\pm k}}{k!}=\frac{e^{\sqrt{5}} \phi^n-(1-\phi)^n}{\sqrt{5}\, \exp(\phi^{\mp 1})},$$ with $\phi$ the golden ratio (the positive solution of $1+1/\phi=\phi$). So the ratio of the $\pm$ series equals $e^{\phi-1/\phi}=e$, and the entire $n$-dependence drops out of the ratio.

The identity also holds for $n\in\mathbb{R}$, with the usual generalisation $F_z=(\phi^z-\phi^{-z}\cos\pi z)/\sqrt{5}$ of the Fibonacci numbers to real argument $z$.

More generally, $$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$ which is equivalent to Will Sawin's identity.

Similarly, $$e^x\sum_{k=0}^\infty F_{n+k}\frac{x^k}{k!}= \sum_{k=0}^\infty F_{n+2k}\frac{x^k}{k!},\tag{2}$$ and more generally, from the formula $$(-1)^q F_{p-q}+F_{\!q}\,\phi^p = F_{\!p}\,\phi^q,$$ where $\phi = (1+\sqrt5)/2$, and the analogous formula with $\phi$ replaced by $(1-\sqrt5)/2$, we get for any integers $p$, $q$, and $n$, $$ e^{(-1)^q F_{p-q}\ x}\sum_{k=0}^\infty F_{q}^k F_{n+pk}\frac{x^k}{k!} = \sum_{k=0}^\infty F_p^k F_{n+qk} \frac{x^k}{k!}. $$ The cases $p=1, q=-1$ and $p=1,q=2$ give $(1)$ and $(2)$.

Related identities (also involving Lucas numbers) can be found in L. Carlitz and H. H. Ferns, Some Fibonacci and Lucas Identities, Fibonacci Quarterly 8 (1970), 61–73.