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 $n\in\mathbb{Z}$.
> 
> 
> 



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


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


