Identity for $e$ in terms of the Fibonacci sequence. The following identity appears in Martin Gardner's paper, "Dr. Matrix on Little Known Fibonacci Curiosities:

$$e = \frac{1 + 1 + \frac{2}{2!} + \frac{3}{3!} + \frac{5}{4!} + \frac{8}{5!} + \frac{13}{6!} + \frac{21}{7!} + \frac{34}{8!} + \frac{55}{9!} + \cdots}{1 + 0 + \frac{1}{2!} + \frac{1}{3!} + \frac{2}{4!} + \frac{3}{5!} + \frac{5}{6!} + \frac{8}{7!} + \frac{13}{8!} + \frac{21}{9!} + \cdots} $$

How can we prove this?
 A: The correct form of this expression seems to be:
$$e=\frac{1+1+\frac{2}{2!}+\frac{3}{3!}+\frac{5}{4!}+\frac{8}{5!}+\frac{13}{6!}+\ldots}{1-0+\frac{1}{2!}-\frac{1}{3!}+\frac{2}{4!}-\frac{3}{5!}+\frac{5}{6!}-\ldots} = \frac{\sum\limits_{k=0}^\infty \frac{F_{k+1}}{k!}}{\sum\limits_{k=0}^\infty (-1)^k\frac{F_{k-1}}{k!}}$$
There are two useful observations that can be used to prove this statement:


*

*$\sum\limits_{k=0}^\infty \frac{x^k}{k!}=e^x$ (the definition of $e^x$)

*$F_n=\frac{\varphi^n-\psi^n}{{\varphi-\psi}}$, where $\varphi=\frac{1+\sqrt{5}}{2}$ and $\psi=\frac{1-\sqrt{5}}{2}$ (the Euler-Binet formula)


Now we are all set for the proof!
$$\begin{eqnarray}
\sum\limits_{k=0}^\infty \frac{F_{k+1}}{k!} & = & \frac{1}{\varphi-\psi}\sum\limits_{k=0}^\infty \frac{\varphi^{k+1}-\psi^{k+1}}{k!} \\
& = & \frac{1}{\varphi-\psi}\left(\varphi\sum\limits_{k=0}^\infty \frac{\varphi^k}{k!}-\psi\sum\limits_{k=0}^\infty \frac{\psi^k}{k!}\right) \\
& = & \frac{1}{\varphi-\psi}\left(\varphi e^\varphi - \psi e^\psi\right) \\
\end{eqnarray}$$
$$\begin{eqnarray}
\sum\limits_{k=0}^\infty (-1)^k\frac{F_{k-1}}{k!} & = & \frac{1}{\varphi-\psi}\sum\limits_{k=0}^\infty \frac{(-1)^k\varphi^{k-1}-(-1)^k\psi^{k-1}}{k!} \\
& = & \frac{1}{\varphi-\psi}\left(
\frac{1}{\varphi}\sum\limits_{k=0}^\infty \frac{(-\varphi)^k}{k!} -
\frac{1}{\psi}\sum\limits_{k=0}^\infty \frac{(-\psi)^k}{k!}\right) \\
& = & \frac{1}{\varphi-\psi}\left(\frac{1}{\varphi} e^{-\varphi} - \frac{1}{\psi} e^{-\psi}\right) \\
\end{eqnarray}$$
Dividing these two yields: $$\frac{\sum\limits_{k=0}^\infty \frac{F_{k+1}}{k!}}{\sum\limits_{k=0}^\infty (-1)^k\frac{F_{k-1}}{k!}} = \frac{\varphi e^\varphi - \psi e^\psi}{ \frac{1}{\varphi} e^{-\varphi} - \frac{1}{\psi} e^{-\psi} }$$
Since $\varphi + \psi = 1$ and $\varphi\psi = -1$, we this can be simplified as 
$$\frac{\varphi e^\varphi - \psi e^\psi}{ \frac{1}{\varphi} e^{-\varphi} - \frac{1}{\psi} e^{-\psi} } = 
\frac{\varphi e^\varphi - \psi e^\psi}{ e^{-1}\left(\varphi e^{\varphi} - \psi e^{\psi}\right)} = e$$
Q.E.D.
