If $x_{n+1}=x_n+x_n^{-m}, x_1>0, m \ge 2 $ then $x_{n}^{m+1} = (m+1)(n+\ln(n))+O_m(1) $ This is a generalization of
Divergence of a sequence proof
Show that
if
$x_{n+1}=x_n+\dfrac{1}{x_n^m},
x_1>0, m \ge 2
$
then
$x_{n}^{m+1}
= (m+1)(n+\ln(n))+O_m(1)
$.
(Note:
The notation $O_m(...)$
means that the
constant implied by the
big-oh
depends on $m$.)
I can show that
$x_{n}^{m+1}
= (m+1)n+O_m(\ln(n))
$
for integer $m$,
and I am pretty sure that
in this case
I can show that
$x_{n}^{m+1}
= (m+1)(n+\ln(n))+O_m(1)
$.
I don't have an
explicit expression for the
$O_m(1)$
and I don't have a proof
for real $m \ge 2$.
 A: The coefficient of $\log n$ turns out to be $\color{blue}{m/2}$, not $m+1$. Also, the "$O_m(1)$" depends on $x_1$ too.
We assume $m>-1$, and use the binomial series (BS) and the Stolz–Cesàro theorem (SCT).
Let $y_n=x_n^{m+1}$; then $y_{n+1}=y_n(1+1/y_n)^{m+1}$. Clearly $y_n\to\infty$ [as $n\to\infty$]. By (BS), $$y_{n+1}=y_n\big(1+(m+1)/y_n+o(1/y_n)\big)\implies y_{n+1}-y_n\to m+1,$$ and (SCT) gives $y_n/n\to m+1$ (as we already know from the OP).
Now we let $y_n=(m+1)n+z_n$ with $z_n=o(n)$, and we put this into the next-order (BS) $$y_{n+1}=y_n\left(1+\frac{m+1}{y_n}+\frac{m(m+1)}{2y_n^2}+o(y_n^{-2})\right),$$ which results in $$z_{n+1}-z_n=\frac{m(m+1)}{2(m+1)n+2z_n}+o(n^{-1})=\frac{m}{2n}+o(n^{-1}),$$ and this time (SCT) yields $z_n/\log n\to m/2$ (as claimed at the beginning).
Similarly, if we let $z_n=(m/2)\log n+t_n$, we get $t_{n+1}-t_n=O(n^{-2}\log n)$, hence $$\lim_{n\to\infty}t_n=\lim_{n\to\infty}\left(x_n^{m+1}-(m+1)n-\frac{m}{2}\log n\right)$$ exists (and is finite; but I wouldn't expect it to have a closed form).
