How do I calculate: $$\lim_{n\to \infty} \frac 1 {n^{2020}} \sum_{k=1}^n k^{2019}$$
I think the answer would be the coeffecient of $n^{2020}$ in the expansion of the summation as that would be the largest term. However, I can't find a way to get this coeffecient.
In general, $$ \lim_{n\to \infty} \frac 1 {n} \sum_{k=a}^b f\left(\frac{k}{n}\right)=\int_{\lim_{n \to \infty}a/n}^{\lim_{ n \to \infty}b/n} f(x) \mathrm{d}x$$ For this particular question, $$\lim_{n\to \infty} \frac 1 {n^{2020}} \sum_{k=1}^n k^{2019}$$ $$=\lim_{n\to \infty} \frac 1 {n} \sum_{k=1}^n \left(\frac{k}{n} \right)^{2019} $$ $$=\int_0^1 x^{2019} \mathrm{d}x = \dfrac{1}{2020}$$ Alternatively, as you mentioned in your post, the answer would indeed be the coefficient of $x^{2020}$ in the summation.
You must be aware of the formulae: $$ \sum_{k = 1}^n k = \frac{n(n+1)}{2} = \frac{n^2}{2} + \frac{n}{2}$$ $$ \sum_{k = 1}^n k^2 = \frac{n(n+1)(2n+1)}{6} = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$$ $$ \sum_{k = 1}^n k^3 = \frac{n^2(n+1)^2}{4} = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}$$ Do you notice something? The coefficient of $n^{r+1}$ in the summation of $\sum_{k = 1}^n k^r$ is equal to $\dfrac{1}{r+1}$ $$ \sum_{k = 1}^n k^{2019} = \frac{n^{2020}}{2020} + an^{2019} + bn^{2018}+\cdots$$ $$ \lim_{n\to \infty} \frac 1 {n^{2020}} \sum_{k=1}^n k^{2019} = \lim_{n\to \infty}\frac{1}{2020} + \frac{a}{n}+ \frac{b}{n^2}+ \cdots = \frac{1}{2020}$$
