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? 

With $y_n=nx-n$, we have the recurrence
$$ y_{n+1} = n(\sqrt[n]{1+y_n}-1)$$
Note that for $h>0$, we have $(1+h)^n>1+nh$, hence  $\sqrt[n]{1+h}<1+\frac hn$, which makes $\{y_n\}$ a decreasing sequence, bounded from below by $0$, hence convergent. Let $L$ be the limit.
With $f_n(t):=\exp(t\ln(1+y_n))$ note that
$$ y_{n+1}=\frac{f_n(1/n)-f_n(0)}{1/n}=f_n'(\tau_n)$$
with $0<\tau_n<\frac1n$. Since $f_n'(t)=\ln(1+y_n)f_n(t)$, this gives us
$$ y_{n+1}=\ln(1+y_n)\exp(\tau_n\ln(1+ y_n)).$$
The right hand side is continuous as a function in $\tau$ and $y$, hence in the limit, when $\tau_n\to 0$ and $y_n\to L$, we find
$$ L=\ln(1+L)\cdot 1$$
which has the only solution $$L=0.$$
