Prove $\left[\ln(n + 1)\right]^p - \left[\ln(n)\right]^p \to 0$ as $n \to \infty$, $p \geq 1$. Problem: I am trying to prove that for each real number $p \geq 1$, that the sequence $\left(\left[\ln(n + 1)\right]^p - \left[\ln(n)\right]^p \right)_{n \geq 1}$ converges to $0$.
I can do it for $p = 1$ using the continuity of the natural logarithm, but I cannot evaluate the limit for $p > 1$. By plotting some test cases on Desmos, I am pretty sure that the result holds for each $p > 1$ as well.
Is it possible to use the continuity of the map $f: \mathbb{R} \to \mathbb{R}; x \to x^p$ and the limit in the case $p = 1$ to prove the general result? I have tried this, but I have made practically no progress.

The start of a proof:
Suppose $\varepsilon > 0$. Then we need to show that there exists $k \in \mathbb{Z}^+$ such that for each integer $n \geq k$, $\left|\left[\ln(n+1)\right]^p - \left[\ln(n)\right]^p\right| < \varepsilon$. I can't go any further than this at the moment.
 A: Let $p >1$. By L'Hopital's Rule you can see that $\frac {\ln (n+1)} {n^{1/(p-1)}} \to 0$. Now by MVT $[\ln (n+1)]^{p}-[\ln n ]^{p}= px^{p-1} [\ln (n+1)-\ln n]$ for some $x$ between $\ln n$ and $\ln (n+1)$. This gives $[\ln (n+1)]^{p}-[\ln n ]^{p} \leq p(\ln (n+1))^{p-1} \ln (1+\frac 1 n)$. Now use the fact that $\ln (1+\frac  1 n) \leq \frac  1 n$ to finish the proof.
A: You can directly use the series expansion at $n=\infty$
$$\log^p(n+1)\\=\left[\log(n) + \frac{1}{n} - \frac{1}{2 n^2} + \frac{1}{3 n^3} - \frac{1}{4 n^4}....\right]^p $$
from which you can easily complete the proof.
A: we always have $lim_{t\rightarrow \infty}ln(t)^{p-1}/t=0$  for all $p>1$,and note $ln(x+1)^p-ln(x)^p$ can be written as:$p\int_{x}^{x+1} ln(t)^{p-1}/t\ dt$, since $lim_{t\rightarrow \infty}ln(t)^{p-1}/t $ is zero,$lim_{x \rightarrow \infty}ln(x+1)^p-ln(x)^p$ would be zero as well.
A: $$\log(n+1)=\log(n(1+\frac1n))=\log n+\log(1+\frac1n)=\log n+\frac1n+o(\frac1n)$$
$$\log(n+1)^p=(\log n+\frac1n+o(\frac1n))^p =$$
$$\log^pn(1+\frac1{n\log n}+o(\frac1{n\log n}))^p=\log^pn(1+\frac p{n\log n}+o(\frac1{n\log n}))$$
$$\log^p(n+1)-\log^pn=\log^pn(\frac p{n\log n}+o(\frac1{n\log n}))$$
