How to determine the limit of this sequence? I was wondering how to determine the limit of $ (n^p - (\frac{n^2}{n+1})^p)_{n\in \mathbb{N}}$ with $p>0$, as $n \to \infty$?
For example, when $p=1$, the sequence is $ (\frac{n}{n+1})_{n\in \mathbb{N}}$, so its limit is $1$.
But I am not sure how to decide it when $p \neq 1$. Thanks in advance!

By Lagrange's theorem, there is a $\xi_n\in(0,1)$ for each $n\in\mathbb N$ such that $(n+1)^p-n^p=p(n+\xi_n)^{p-1}$, so we have:
$$n^p -\left(\frac{n^2}{n+1}\right)^p = \frac{n^p((n+1)^p-n^p)}{(n+1)^p}=\frac{n^pp(n+\xi_n)^{p-1}}{(n+1)^p}=p\left(\frac{n}{n+1}\right)^p(n+\xi_n)^{p-1}.$$
Now, for any $p\in(0,\infty)$, the expression $\left(\frac{n}{n+1}\right)^p$ will converge to $1$. For $p>1$, the expression $(n+\xi_n)^{p-1}$ will converge to $+\infty$ and for $p\in(0,1)$ will converge to $0$. So your original expression will also converge to $+\infty$ and $0$, respectively, in these two cases.
