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. 

Note that
$$
\begin{align}
\frac{x}{1-x}
&=\sum_{n=1}^\infty x^n\\
&=\sum_{n=1}^\infty\sum_{k=1}^n\frac{a_k}{(n-k+1)!}x^n\\
&=\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{a_k}{(n-k+1)!}x^n\\
&=\sum_{k=1}^\infty\sum_{n=0}^\infty\frac{a_k}{(n+1)!}x^{n+k}\\
&=\frac{e^x-1}{x}\sum_{k=1}^\infty a_kx^k\tag{1}
\end{align}
$$
Therefore,
$$
\sum_{k=1}^\infty a_kx^k=\frac{x^2}{(e^x-1)(1-x)}\tag{2}
$$
If $a_k$ limit to some $b$, then as $x\to1$ we would have
$$
\begin{align}
b
&=\lim_{x\to1^-}\frac{\displaystyle\sum_{k=1}^\infty a_kx^k}{\displaystyle\sum_{n=0}^\infty x^k}\\
&=\lim_{x\to1^-}\frac{x^2}{e^x-1}\\[9pt]
&=\frac1{e-1}\tag{3}
\end{align}
$$
Now that we have an idea of what the limit would be, let's try to prove it.
From the defining formula,
$$
a_n=1-\sum_{k=1}^\infty\frac{a_{n-k}}{(k+1)!}\tag{4}
$$
where we define $a_k=0$ for $k\le0$.
Thus, if we let $c_n=a_n-\frac1{e-1}$, then for $n\gt1$, we have
$$
c_n=-\sum_{k=1}^\infty\frac{c_{n-k}}{(k+1)!}\tag{5}
$$
Note that if $|c_k|\lt c\,(4/5)^k$ for $k\lt n$ then
$$
\begin{align}
|c_n|
&\le\sum_{k=1}^\infty\frac{c\,(4/5)^{n-k}}{(k+1)!}\\
&=c\,(4/5)^n\sum_{k=1}^\infty\frac{(4/5)^{-k}}{(k+1)!}\\
&=c\,(4/5)^n\frac45\left(e^{5/4}-1-\frac54\right)\\[9pt]
&\le c\,(4/5)^n\tag{6}
\end{align}
$$
Since $c_1=\frac{e-2}{e-1}$ and $c_n=-\frac1{e-1}$ for $n\le0$, we can use $c=\frac1{e-1}$ in $(6)$. Therefore,
$$
\begin{align}
\left|\,a_n-\frac1{e-1}\,\right|
&=|c_n|\\
&\le\frac{(4/5)^n}{e-1}\tag{7}
\end{align}
$$
Therefore,
$$
\lim_{n\to\infty}a_n=\frac1{e-1}\tag{8}
$$
