Sequence and series problem How do I show that the sequence $(x_n)$ defined by $$x_ {n+1} = \left(1-\frac{1}{n}\right) ^2 x_n + \frac{1}{n}, \forall \,n \in \Bbb{N}-\left\{0\right\}$$ converges? and to what limit?
 A: Hint. Just multiply by $n^2$ to obtain
$$
n^2x_{n+1}=(n-1)^2x_{n}+n, \quad n=1,2,3,\ldots \tag1
$$ set $u_n:=(n-1)^2x_{n}+C$, for any constant $C$, thus you obtain
$$
u_{n+1}-u_n=n \tag2
$$ this is a telescoping sum which leads to the result if you know how to evaluate $\displaystyle \sum_{1}^{N}n.$
Summing $(2)$, from $n=1$ to $n=N$, $N\geq1$, we have
$$
(u_{2}-u_1)+(u_{3}-u_2)+\ldots+(u_{N+1}-u_N)=\frac{N(N+1)}{2} \tag3
$$
on the left hand side, clearly all terms cancel except two of them, we get
$$
u_{N+1}-u_1=\frac{N(N+1)}{2}
$$ or
$$
N^2x_{N+1}-C=\frac{N(N+1)}{2}, \quad x_{N+1}=\frac{C}{N^2}+\frac{N+1}{2N} \tag4
$$ but from the initial identity defining the sequence $\left\{x_n\right\}_{n\geq 1}$, we have $u_2=1$, implying $C=0$, finally we have

$$
x_N=\frac{1}{2(N-1)}+\frac12,\quad N=2,3,4,\ldots, \tag5
$$ 

$u_1$ being any constant, the limit being easy to obtain.
A: Hint: If you substitute $y_n=(n-1)^2x_n$, the recurrence relation becomes:
$$y_{n+1}=y_{n}+n.$$
A: proving by induction we get
$$x_n=\frac{n}{2(n-1)}+\frac{C}{(n-1)^2}$$ thus for $n$ goes to infinity we obtain the limit $\frac{1}{2}$
