How would someone find the limit of the sequence $a_n = \frac{1^k+2^k+...+n^k}{n^{k+1}}, k \in \mathbb{N}$ as $n$ goes to Infinity? Can someone give me maybe a hint where to start?

By considering: $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^1}{n^{2}} = \frac 1 2$$ $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^2}{n^{3}} = \frac 1 3$$ $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^3}{n^{4}} = \frac 1 4$$ Determine if this is true: $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}} = \frac ...

What is the result of the next limit: $$ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} $$ Why (theorem)?

How do we prove that $$\lim_{n \rightarrow \infty} \dfrac{\displaystyle\sum_{r=1}^{n} r^a}{n^{a+1}}=\frac{1}{a+1}$$? This type of terms appear in problems on limits, but I am unable to prove this. Please help me out.

Evaluate $$\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)$$

I've calculated $\lim_{n\to\infty}\dfrac{1^p+2^p+\cdots+n^p}{n^{p+1}}=\dfrac1{p+1}$ where $p\in\mathbb{N}$ fixed. I feel it should help me get this one $\lim_{n\to\infty}\left(\dfrac{1^p+2^p+\cdots+n^p}{n^{p}}-\dfrac{n}{p+1}\right)$, but I'm not sure how. Any hints?

I have$$\lim_{n\to \infty} \frac{1^p+2^p+\ldots+n^p}{n^{p+1}}=$$ I managed to simplify it down to $$=\lim_{n\to \infty}\left( \left(\frac1n\right)^p \cdot \frac1n + \left(\frac2n\right)^p \cdot \frac1n + \ldots + \left(\frac{n}{n} \right)^p \cdot \frac1n \right)=$$ $$=\lim_{n\to \infty} \sum_{i=1...