Limit of $a(k)$ in $ \sum_{k=1}^n \frac{a_k}{(n+1-k)!} = 1 $ For n = 1, 2, 3 ... (natural number)
$ \sum_{k=1}^n \frac{a_k}{(n+1-k)!} = 1 $
$ a_1 = 1, \ a_2 = \frac{1}{2}, \ a_3 = \frac{7}{12} \cdots $
What is the limit of {$ a_k $}
$ \lim_{k \to \infty} a_k $ = ?
I have no idea where to start. 

Put $\displaystyle f(x)=\sum_{k\geq 1} a_k x^{k-1}$, $\displaystyle g(x)=\sum_{l\geq 0} \frac{x^l}{(l+1)!}=\frac{\exp(x)-1}{x}$, and $\displaystyle h(x)=f(x)g(x)=\sum_{n\geq 0}c_n x^n$. The coefficient $c_n$ is equal to $1$ if $n=0$, and for $n\geq 1$:
$$c_n=\sum_{k-1+l=n, k\geq 1, l\geq 0}\frac{a_k}{(l+1)!}=\sum_{k=1}^{n}\frac{a_k}{(n+1-k)!}=1$$
Hence:
$$(\frac{\exp(x)-1}{x})f(x)=\sum_{n\geq 0}x^n=\frac{1}{1-x}$$
and $\displaystyle f(x)=\frac{x}{(1-x)(\exp(x)-1)}$. Put now $\displaystyle r(x)=(1-x)f(x)=\sum_{m\geq 0}d_m x^m=\frac{x}{\exp(x)-1}$. We have $d_0=a_1=1$, and for $m\geq 1$, $d_m=a_{m+1}-a_{m}$.
The radius of convergence of the power series $r(x)$ is $\geq 2\pi$, because it is regular at $x=0$, and the function $\exp(z)-1$ is non  zero for $z\in \mathbb{C}$, $z\not = 0$ and $|z|<2\pi$. Hence we can put $x=1$ in the formula $\displaystyle r(x)=\frac{x}{\exp(x)-1}$, we have hence $\displaystyle r(1)=\frac{1}{e-1}$. 
Now $r(1)$ is the limit of $d_0+\cdots d_{n-1}=t_n$ if $n\to +\infty$; but one can  easily see that $t_n=a_n$, so $\displaystyle a_n\to \frac{1}{e-1}$. 
