Interesting Limit (28)

Here I used the fact that the result of $\displaystyle \sum_{k=1}^n k^m$ can be expressed as a polynomial in $n$ of degree $m+1$ and leading coefficient $\frac 1{m+1}$ . Thus, it suffices to compute

$\displaystyle \lim_{n\to\infty} \frac{\ln\Big(\frac 1{2021}n^{2021}\Big) } {\ln n} =$

$\displaystyle =\lim_{n\to\infty} \frac{2021\ln n - \ln 2021} {\ln n} =$

$\displaystyle =\lim_{n\to\infty} \bigg(2021-\frac{\ln 2021} {\ln n}\bigg)=2021$