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? 
 A: Since $(1+x)^n\geqslant nx+1$ we obtain that $x_n\geqslant x_{n+1}$. As the sequence is positive and decreasing, it must converge. Call this limit $\ell$. Consider the non-negative functions $$f_n(x)=\frac{\log(1+nx)}n$$
They have the property that $$\log(1+x_{n+1})=f_n(x_n)$$
Since $x_n$ is decreasing, and since the $f_n$ are decreasing, meaning that $f_{n+1}\leq f_n$, yet each one of them is increasing, meaning $f_n(x)\leq f_n(y)$ if $x\leq y$, we have $$f_{n+1}(x_{n+1})\leq f_n(x_{n+1})\leq f_n(x_n) $$
The limit thus exists. We would like to argue that $f_n(x_n)\to 0$. Since $x_n$ decreases and is non-negative, we can work inside a compact interval $[0,M]$. In this interval, the continuous $f_n$ converge monotonically to $0$, thus by Dini's theorem, they converge uniformly. Thus, $\{f_n\}_{n\geqslant 1}$ is an equicontinuous family, whence $$\lim_{n\to\infty}f_n(x_n)=\lim_{n\to\infty}f_n(\ell)=0$$
But then $\log(1+\ell)=0\implies \ell =0$. 
Hint  Let ${\left( {1 + {x_{n + 1}}} \right)^n} = 1 + n{x_n} = {y_n}$. Since $x_n\to 0$ we may expand $$\log \left( {1 + {x_{n + 1}}} \right) = {x_{n + 1}} + {x_{n + 1}}O\left( {{x_{n + 1}}} \right)$$ for sufficiently large $n$, so if we let $y_n-1=\omega_n$; $$\log \left( 1+\omega_n \right) = \frac{n}{{n + 1}} \omega_{n+1} + n{x_{n + 1}}O\left( {{x_{n + 1}}} \right)$$
Thus if the limit exists, it must be $0$.

Let $\mathscr F=\{f_i\}_{i\in I}$ be a family of functions $f_i:A\to\Bbb R$. We say $\mathscr F$ is equicontinuous on $A$ if for each $\epsilon>0$ there exists $\delta >0$ such that for each $x,y\in A$ and $f_i\in\mathscr F$ $$|x-y|<\delta\implies |f_i(x)-f_i(y)|<\epsilon$$
Note every function in an equicontinuous family of functions is automatically uniformly continuous, for example.
A: 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 $n\to\infty$.
A: 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 $$L=0.$$
