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?
It is clear that $ x_n&gt; \frac{1}{n}.$

Using mathematical induction is easy to see that $ x_n &lt; \frac{1}{n-4} $ for $n\geq6.$

Proof by mathematical induction:

$x_6=\frac{199}{450}&lt;\frac{1}{2}$

$x_{n+1}= \frac{n+1}{n^2} + \frac{x_n}{n}&lt; \frac{n+1}{n^2}+\frac{1}{n(n-4)}=\frac{n^2-2n-4}{n^2(n-4)}. $

Because $\frac{n^2-2n-4}{n^2(n-4)}&lt;\frac{1}{n-3}&lt;=&gt;-n^2+2n+12&lt;0$ assertion is proved.

It follows that for $n\geq6$  $$\frac{1}{n}&lt;x_n &lt; \frac{1}{n-4} $$

and consequently $\ lim_{n\to\infty} nx_n=1.$
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 $1$.