Find the limit of following sequence $\lim\limits_{n \to inf}\frac{n×1^r+(n-1)×2^r+...+1×n^r}{n^{r+2}}$
Ok so the only idea I got was to use Stholz theorem and then use $1^r+2^r+...+n^r$ ~ $
\frac{1}{(r+1)} (n^{r+1})$ somehow but that didn't lead anywhere (or at least I didn't see it)
 A: I don't know about the Stolz theorem, but the numerator is the Riemann sum for
$$\int_0^1 (1-x)(1+ n x)^r d x.$$ The integral is not hard to evaluate in closed form, so the limit of the sequence is $0.$
A: Igor Rivin's idea is great, but the limit is not $0$. We need to assume that $r\neq -1$ and $r\neq -2$. The general term of the given sequence is
\begin{align*}
 S_n(r)   &  = \frac{1}{n^{r+2}} \sum_{k=1}^{n} (n-k+1)k^r \\ 
  & = \frac{n+1}{n^{r+2}} \sum_{k=1}^{n} k^r - \frac{1}{n^{r+2}} \sum_{k=1}^{n} k^{r+1} \\
  & = \frac{n+1}{n} \frac{1}{n} \sum_{k=1}^{n} \left(\frac{k}{n}\right)^{r} - \frac{1}{n} \sum_{k=1}^{n} \left(\frac{k}{n}\right)^{r+1}. 
\end{align*}
We have that
\begin{equation*}
\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \left(\frac{k}{n}\right)^{r} = \frac{1}{r+1}, 
\end{equation*}
since the terms of this seqence are Riemann sums of $\int_{0}^{1} x^r dx = \frac{1}{r+1}$ and
\begin{equation*}
\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \left(\frac{k}{n}\right)^{r+1} = \frac{1}{r+2}, 
\end{equation*}
since the terms of this sequence are Riemann sums of $\int_{0}^{1} x^{r+1} dx = \frac{1}{r+2}$.
Therefore
\begin{equation*}
\lim_{n\to\infty} S_n(r) = \frac{1}{r+1} - \frac{1}{r+2} = \frac{1}{(r+1)(r+2)}. 
\end{equation*}
For $r=-1$ we have
\begin{equation*}
 S_n(-1)  = \frac{n+1}{n} H_n - 1. 
\end{equation*}
Therefore,
\begin{equation*}
\lim_{n\to\infty} S_n(-1)  = \infty.  
\end{equation*}
For $r=-2$ we have
\begin{equation*}
 S_n(-2)  = (n+1) H_{n,2} - H_n. 
\end{equation*}
Since,
\begin{equation*}
\lim_{n\to\infty} H_{n,2} =\frac{\pi^2}{6},  
\end{equation*}
we have
\begin{equation*}
\lim_{n\to\infty} S_n(-2)  = \infty.  
\end{equation*}
