Showing that $\lim_{n \rightarrow \infty}\frac{1^k+2^k+...+n^k}{n^{k+1}}=\frac{1}{k+1}$ using integral

Question

How can I show (using upper-, lower-, Riemann-sums) that $$\lim_{n \rightarrow \infty}\frac{1^k+2^k+...+n^k}{n^{k+1}}=\frac{1}{k+1}$$ with $k, n \in \mathbb{N}$?

The hint I received is that $\frac{x^{k+1}}{k+1}$ is a antiderivative of $x^k$ and $\int_a^bf(x)\mathrm{d}x=F(b)-F(a)$.

Answer 1

Since $f(x):=x^{k}$ is increasing over $[0,1]$ then, with respect to the partition $x_j=j/n$ for $j=0,\dots,n$ the upper-Riemann-sum is equal to $$S_n=\frac{1}{n}\sum_{j=1}^n f(x_j)=\frac{1}{n}\sum_{j=1}^n (j/n)^{k},$$ and the lower-Riemann-sum is $$s_n=\frac{1}{n}\sum_{j=0}^{n-1} f(x_j)=\frac{1}{n}\sum_{j=0}^{n-1} (j/n)^{k}.$$ Hence $$S_n-\frac{1}{n}=s_n\leq \int_0^1 f(x) dx\leq S_n\implies 0\leq S_n-\int_0^1 f(x) dx \leq \frac{1}{n}$$ and therefore, by th Squeeze Theorem, as $n$ goes to infinity, $$\frac{1^k+2^k+...+n^k}{n^{k+1}}=S_n\to \int_0^1 f(x) dx=\int_0^1 x^k dx=\left[\frac{x^{k+1}}{k+1}\right]_0^1=\frac{1}{k+1}.$$

Answer 2

Partition on $[0,1]$ that $P_{n}=\{1/n,2/n,...,n/n\}$, and $\lim_{n\rightarrow\infty}S(f,P_{n})=\lim_{n\rightarrow\infty}\dfrac{1}{n}\dfrac{1^{k}+\cdots+n^{k}}{n^{k}}=\displaystyle\int_{0}^{1}x^{k}dx=\dfrac{1}{k+1}$, where $f(x)=x^{k}$, $x\in[0,1]$.