<s>[INST] 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?
 [/INST] $$x_n=n(x_{n+1}-\frac{n+1}{n^2})$$
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
$$\lim_{n\to \infty} x_n = 0$$
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