Approximating sums of powers of integers: why does $\sum_{i=1}^n i^r \div \frac{n^{r+1}}{r+1} \to 1$ as $n \to \infty$?

Question

I know there are exact formulas for sums of integer powers of integers ($\sum_{i=1}^n i^r$ with $r \in \mathbb{N}$), but I was interested in approximating them. One way occured to me through a geometrical argument. If you arrange $1,2,3,...,n-1,n$ as a right-angled triangle, they fill up about half of the square $n^2$. Similarly if you arrange $1^2,2^2,3^2,...,(n-1)^2,n^2$ as a 'right-angled' pyramid, they fill up about a third of the cube $n^3$. This suggests the general approximation: $$\sum_{i=1}^n i^r \approx \frac{n^{r+1}}{r+1}$$ and indeed I've found numerically that $$\lim_{n\to\infty}{\frac{\sum_{i=1}^n i^r}{\frac{n^{r+1}}{r+1}}} \to 1.$$How can I prove this?

Answer 1

One can show by induction on $r$ that $\sum_{i=1}^n i^r$ is a polynomial funtion in $n$ of degree $d=r+1$. Then noting that $(n+1)^d=n^d+d\cdot n^{d-1}+{d\choose 2}n^{d-2}+\ldots$ you get the asymptotic result.