Compute $\lim_{n\to\infty} nx_n$ Let $(x_n)_{n\ge2}$, $x_2>0$, that satisfies recurrence $x_{n+1}=\sqrt[n]{1+n x_n}-1, n\ge 2$. Compute $\lim_{n\to\infty} nx_n$.
It's clear that $x_n\to 0$, and probably Stolz theorem would be helpful. Is it really necessary to use this theorem? 

Consider the function $$f_n(t)=\Bigl(1+\frac{t}{n+1}\Bigr)^n,\qquad t\ge0, \quad n=2,3,\ldots$$
Clearly (convexity, value and derivative at $t=0$, growth), there is a $t_n>0$ so that $f_n(t)<1+t$ when $0<t<t_n$, and $f_n(t)>1+t$ when $t>t_n$. Now
$$f_n(t)\ge 1+\frac{n}{n+1}t+\frac{n(n-1)}{2(n+1)^2}t^2=1+\frac{nt}{n+1}\Bigl(1+\frac{n-1}{2(n+1)}t\Bigr)>1+t$$
provided $$\frac{n-1}{2(n+1)}t>\frac1n,$$
and it follows that $$\limsup_{n\to\infty}nt_n\le2.$$
Whenever $nx_n>t_n$ we have $f_n(nx_n)>1+nx_n$, that is
$$1+\frac{nx_n}{n+1}>\sqrt[n]{1+nx_n}=1+x_{n+1}$$
and so $(n+1)x_{n+1}<nx_n$.
Even when this is not the case, $x_{n+1}<x_n$, so that $(n+1)x_{n+1}<nx_n+x_n$.
It is possible that $nx_n>t_n$ for all large $n$. If so, the sequence $(nx_n)$ is decreasing, so it has a limit, which must be zero (see the answer by Peter Tamaroff).
If $nx_n\le t_n$ for arbitrarily large $n$, then for such $n$, $(n+1)x_{n+1}<t_n+x_n$,
which goes to zero as $n\to\infty$ (see the comments for an argument that $x_n\to0$). So in any case, $nx_n\to0$ as $n\to\infty$.
