Finding limit of a recursively defined sequence Let $(x_{n})_{n\geq1}$ be a sequence defined by:
$x_1=1$ and $x_n=n(x_{n+1}-\frac{n+1}{n^2})$. Calculate $\lim_{x\rightarrow \infty }nx_n$.
We can write $x_n=n(x_{n+1}-\frac{n+1}{n^2})$ as $(n+1) x_{n+1} = \frac{n + 1}{n^2}(n x_n + (n+1))$, and by the substitution $y_n=nx_n$ we obtain: $y_{n+1} =  \frac{n + 1}{n^2}y_n + \left(1 + \frac{1}{n}\right)^2$. How to go on?
 A: It is clear that $ x_n> \frac{1}{n}.$
Using mathematical induction is easy to see that $ x_n < \frac{1}{n-4} $ for $n\geq6.$
Proof by mathematical induction:
$x_6=\frac{199}{450}<\frac{1}{2}$
$x_{n+1}= \frac{n+1}{n^2} + \frac{x_n}{n}< \frac{n+1}{n^2}+\frac{1}{n(n-4)}=\frac{n^2-2n-4}{n^2(n-4)}. $
Because $\frac{n^2-2n-4}{n^2(n-4)}<\frac{1}{n-3}<=>-n^2+2n+12<0$ assertion is proved.
It follows that for $n\geq6$  $$\frac{1}{n}<x_n < \frac{1}{n-4} $$
and consequently $\ lim_{n\to\infty} nx_n=1.$ 
A: All right. You wanted $nx_n$.
so
$y_n = nx_n$
$$y_n=n^2(\frac{y_{n+1}}{n+1}-\frac{n+1}{n^2}) = y_{n+1}\frac{n}{n+1} - n  - 1)$$
$$y_{n+1}=\frac{n+1}{n}(y_n + n  + 1)$$
then you'll step into solution of $y_{n+1}=y_n$ and
$$y_{n}=\frac{n+1}{n}(y_n + n  + 1)$$
$$y_{n}\frac{1}{n}=\frac{n+1}{n}(n  + 1)$$
$$y_{n}=(n+1)^2$$
this means, that there is no such a limit. (have you tried to list first few points of this serie?)
A: If the sequence converges, then when $n\to \infty$, we can set $c=x_n=x_{n+1}$, and the limit $c$ satisfies
$$c=n\left(c-\frac{n+1}{n^2}\right)$$
from which $c$ is solved to be $1/n$.
The limit $\lim_{n\to \infty}nx_n$ is thus $1$.
A: $$x_n=n(x_{n+1}-\frac{n+1}{n^2})$$
you may rewrite that to some nice form:
$$x_{n+1} = \frac{n+1}{n^2} + \frac{x_n}{n}$$
with $n\to \infty$ you'll end up by solving equation: $x_{n+1} = x_n$, which you may substitute to the $$x_n = \frac{n+1}{n^2} + \frac{x_n}{n}$$
this may be rearranged to:
$$x_n(1-\frac{1}{n}) = \frac{n+1}{n^2}$$
now make the limit of both sides: $$\lim_{n\to \infty}(1-\frac{1}{n}) = 1$$ and $$\lim_{n\to \infty}\frac{n+1}{n^2} = 0$$ so
$$\lim_{n\to \infty} x_n = 0$$
