Limit of $u_{n+1}=-\frac{1}{e}+(n+1)u_n, u_0=1-1/e$ Suppose that the sequence $\{u_n\}_{n\ge 0}$ is given by the recurrence relation
$u_{n+1}=-\frac{1}{e}+ (n+1)u_n, u_0=1-1/e$. 
Is there a direct way to deduce its limit? The term $(n+1)u_n$ is what bothers me.
 A: The limit is, in fact, zero.
We first prove by induction on $n$ that the following form holds:
$$
a_n
=n! \left(1-\frac{1}{e}\sum_{i=0}^n \frac{1}{i!}\right) \tag{1}$$
By direct substitution this matches the initial condition $a_0=1-\frac{1}{e}$.
Now suppose $(1)$ holds for $n=k$. 
Thereby
$$
a_k
=k!\left(1-\frac{1}{e}\sum_{i=0}^{k} \frac{1}{i!}\right)
$$
and
\begin{align}
a_{k+1}
& =-\frac{1}{e}+(k+1)! \left(1-\frac{1}{e}\sum_{i=0}^k\frac{1}{i!}\right) \\
& = (k+1)!\left(1-\frac{1}{e}\left(\sum_{i=0}^k \frac{1}{i!}\right) + \frac{1}{(k + 1)!}\right)
\qquad \left[\frac{1}{e}\to(k+1)!\frac{1}{e} \frac{1}{(k + 1)!}\right] \\
& =(k+1)!\left(1-\frac{1}{e}\sum_{i=0}^{k+1}\frac{1}{i!}\right)
\end{align}
To prove that this goes to zero render
$$
0
=1-\frac{1}{e}\sum_{i=0}^\infty \frac{1}{i!}
$$
and subtract from $(1)$ to obtain
$$
a_n
=n! \frac{1}{e} \sum_{i=n+1}^\infty\frac{1}{i!}.
$$
Taking advantage of the fact that $(n+m)!\ge n!(n+1)^m$ for all $m\ge 1$, this may be bracketed between zero and a geometric series:
$$
0
<a_n
\le \frac{1}{e}\sum_{i=1}^\infty \frac1{(n+1)^i}.
$$
By the standard formula for geometric series the upper bound converges to $\frac{1}{en}\to 0$, forcing $a_n\to 0$ with the Squeeze Theorem.
A: Here is a way
if you don't know the result.
Suppose
$u_{n+1}
=a_n+b_nu_n
$.
We want to convert this
to a telescoping recurrence.
To do this
we want to divide it
by a $B_{n+1}
$
to get it in the form
$\dfrac{u_{n+1}}{B_{n+1}}
=\dfrac{a_n}{B_{n+1}}+\dfrac{b_nu_n}{B_{n+1}}
$
so that
$\dfrac{b_n}{B_{n+1}}
=\dfrac1{B_n}
$.
The recurrence will then be
$\dfrac{u_{n+1}}{B_{n+1}}
=\dfrac{a_n}{B_{n+1}}+\dfrac{u_n}{B_{n}}
$
which telescopes.
For this to hold,
we want
$\dfrac{b_n}{B_{n+1}}
=\dfrac1{B_n}
$
or
$B_{n+1}
=b_nB_n
$.
An obvious way
for this to happen
is to make
$B_n
=\prod_{k=0}^{n-1} b_k
$,
so
$B_{n+1}
=\prod_{k=0}^{n} b_k
$.
In the original problem,
$b_n = n+1$,
so
$B_n
=\prod_{k=0}^{n-1} b_k
=\prod_{k=0}^{n-1} (k+1)
=n!
$.
The recurrence is now
$\dfrac{u_{n+1}}{B_{n+1}}-\dfrac{u_n}{B_{n}}
=\dfrac{a_n}{B_{n+1}}
$
so, summing,
$\sum_{n=0}^{m-1}\dfrac{a_n}{B_{n+1}}
=\sum_{n=0}^{m-1}(\dfrac{u_{n+1}}{B_{n+1}}-\dfrac{u_n}{B_{n}})
=\dfrac{u_{m}}{B_{m}}-\dfrac{u_0}{B_{0}}
$
or
$\dfrac{u_{m}}{B_{m}}
=\dfrac{u_0}{B_{0}}+\sum_{n=0}^{m-1}\dfrac{a_n}{B_{n+1}}
$
so that
$u_m
=u_0B_m+B_m\sum_{n=0}^{m-1}\dfrac{a_n}{B_{n+1}}
$.
In this case,
$b_n = n+1$
so that
$B_n = n!$
and 
$a_n = -\dfrac1{e}$
so that
$\begin{array}\\
u_m
&=m!(1-\dfrac1{e})-\dfrac1{e}m!\sum_{n=0}^{m-1}\dfrac1{(n+1)!}\\
&=m!((1-\dfrac1{e})-\dfrac1{e}\sum_{n=1}^{m}\dfrac1{n!})\\
&=\dfrac{m!}{e}(e-\sum_{n=0}^{m}\dfrac1{n!})\\
&=\dfrac{m!}{e}\sum_{n=m+1}^{\infty}\dfrac1{n!}\\
&=\dfrac{1}{e}\sum_{n=m+1}^{\infty}\dfrac{m!}{n!}\\
&=\dfrac{1}{e}\sum_{n=m+1}^{\infty}\dfrac{1}{\prod_{k=m+1}^{n}k}\\
&\le\dfrac{1}{e}\sum_{n=m+1}^{\infty}\dfrac{1}{\prod_{k=m+1}^{n}(m+1)}\\
&=\dfrac{1}{e}\sum_{n=m+1}^{\infty}\dfrac{1}{(m+1)^{n-m}}\\
&=\dfrac{1}{e}\sum_{n=1}^{\infty}\dfrac{1}{(m+1)^{n}}\\
&=\dfrac{1}{e}\dfrac{\dfrac1{m+1}}{1-\dfrac1{m+1}}\\
&=\dfrac{1}{em}\\
\end{array}
$
A: If we set $u_n:=\dfrac{n!}e\,v_n$,
$$u_{n+1}=-\frac1e+(n+1)u_n$$
turns to
$$v_{n+1}=v_n-\frac1{(n+1)!}$$
and by induction
$$v_n=v_0-\sum_{k=1}^n\frac1{k!}=e-1-\sum_{k=1}^n\frac1{k!}=\sum_{k=n+1}^\infty\frac1{k!}.$$
Finally,
$$\frac{n!\,v_n}e=\frac1e\sum_{k=n+1}^\infty\frac{n!}{k!}=\frac1e\left(\frac1{n+1}+\frac1{(n+1)(n+2)}+\cdots\right)\to0$$
