first question here, so let's hope this goes well. In first-year university Calculus 1, I was learning about the formulas for sums of various powers. For example, $$\sum_{k=1}^n{k}=\frac{n(n+1)}{2}$$ $$\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$$ $$\vdots$$ When used in the context of defining integration as the limit of Riemann sums, these often get used in the form $$\lim_{n\to\infty}\frac{1}{n^{a+1}}\sum_{k=1}^n{k^a},a\in\mathbb{Z}^+\cup\{0\}$$
Taking the limit of the first few formulas yields the following: $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n{k^0}=\lim_{n\to\infty}\frac{1}{n}n=\frac{1}{1}$$ $$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n{k^1}=\lim_{n\to\infty}\frac{1}{n^2}\frac{n(n+1)}{2}=\frac{1}{2}$$ $$\lim_{n\to\infty}\frac{1}{n^3}\sum_{k=1}^n{k^2}=\lim_{n\to\infty}\frac{1}{n^3}\frac{n(n+1)(2n+1)}{6}=\frac{1}{3}$$ $$\lim_{n\to\infty}\frac{1}{n^4}\sum_{k=1}^n{k^3}=\lim_{n\to\infty}\frac{1}{n^4}\frac{n^2(n+1)^2}{4}=\frac{1}{4}$$ $$\lim_{n\to\infty}\frac{1}{n^5}\sum_{k=1}^n{k^4}=\lim_{n\to\infty}\frac{1}{n^5}\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}=\frac{1}{5}$$ $$\vdots$$ $$\lim_{n\to\infty}\frac{1}{n^{a+1}}\sum_{k=1}^n{k^a}=\frac{1}{a+1}$$
Does the general trend in the last equation there hold for all $a$?
