A sum involving Fibonacci numbers, $\sum_{k=1}^\infty F_k/k!$ Let $F_k$ be Fibonacci numbers. I am looking for a closed form of the sum  $\sum_{k=1}^\infty F_k/k!$.

I tried to use Wolfram Alpha, but it is not doing the sum Fibonacci[k]/k! , k=1 to infinity.

Can someone tell what is the problem with WA and what this sum equals to?
A hint:

You can prove for yourself or find in a book a formula of the form
$${\rm fib}(k)=a\lambda^k + b\mu^k$$
with certain constants $a$, $b$, $\lambda$, $\mu$. The requested sum can then be easily written as a sum of two exponentials.
Solve y'' = y' + y both via power series and then by elementary methods (substitution of exponentials) and setting boundary conditions y(0) = 0, and y'(0) = 1. The result is the following identity:

$\sum_{n=1}^\infty \frac{F_n}{n!} x^n = \frac{2}{\sqrt{5}} e^\frac{x}{2} \sinh(\frac{\sqrt{5}}{2}x)$

If you don't feel like solving the differential equation yourself you can easily verify by substituting each side of the above equation into the differential equation and boundary conditions and then remind yourself via the existence and uniqueness theorem that the two solutions must be the same. Meaning they are the same at the point x=1.

Setting x=1 gives 

$\sum_{n=1}^\infty \frac{F_n}{n!} = 2 \sqrt{\frac{e}{5}} \sinh(\frac{\sqrt{5}}{2}) = \frac{e^{\phi_+} - e^{\phi_-}}{\sqrt{5}}$

where $\phi_\pm = \frac{1 \pm \sqrt{5}}{2}$ are the golden ratio and it's congugate. I.e. the two solutions to the quadratic equation $y^2 = y + 1$. (Note that this is the same as the original differential equation with the order of the derivative replaced by the power of y: $y^{(n)} \rightarrow y^n$. Is there a reason for this or is it just an interesting coincidence?)

The numerical value to 1234 decimal places is: 

2.0143227334583157365813462554697591356591114695811241821088403766742128397097006637111011319457016312404491456095258793427009006861303485673596801638909160410193717669927023995171048184677900815088560097082872535199871890392172995661754732113194156588018029328948474213043876788216048899918772417664402571617571139330772358975323915258959445759063347851074795879633549994051398333466312388233898626203203269110709326570399986964305691732913351786602533868094239267790638117143928091599301761319009602970110828764189441445889515565022310301247296449293897072832251096862983584326320448058898294430874162784325966040344781269902152316671151601869355616381547811631999794833349897492383827197553543970970099927748899115338963618621987298237221140308693191390458816782367508072038660426129042190625209844154130995652380767395098326152648319830337117521840226032419661899959469674966107384253745331849405956049149602323282511659984227035172708526550586977824786544373977008821755560614931363394267293413160342932047952358802051132151029521708819143029550619292413657149885241166291353201988752249793933504301075491208793989947801376838994380077603990476851019102426649383266999848219352544560978465856984179911553255421975337456890201316269

A fun exercise is to write a C program to generate this (not as easy as you might assume due to the limits of native C data types).