TITLE: Fibonacci series captures Euler $e=2.718\dots$
QUESTION [35 upvotes]: The Fibonacci recurrence $F_n=F_{n-1}+F_{n-2}$ allows values for all indices $n\in\mathbb{Z}$. There is an almost endless list of properties of these numbers in all sorts of ways. The below question might even be known. Yet, if true, I like to ask for alternative proofs.

Question. Does the following identity hold? We have
  $$\frac{\sum_{k=0}^{\infty}\frac{F_{n+k}}{k!}}{\sum_{k=0}^{\infty}\frac{F_{n-k}}{k!}}=e,$$
  independent of $n\in\mathbb{Z}$.

REPLY [18 votes]: 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.