Can anyone suggest other than Stolz-Cesaro Theorem to solve this? $$\lim _{n \rightarrow \infty} \frac{1}{n^{2020}} \sum_{k=1}^{n}(k+2019)^{2019}$$
My approach:
using Stolz-Cesaro Theorem
let,\begin{aligned}
&a_{n}=\sum_{k=1}^{n}\left(k+2019\right)^{2019}, b_{n}=n^{2020}\\
&a_{n+1}-a_{n}=(n+2020)^{2019}=n^{2019 \ldots}\\
&\begin{array}{l}
b_{n +1}-b_{n}=(n+1)^{2020}-n^{2020} \\
then,
\end{array}\\
&\lim _{n \rightarrow \infty} \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=\frac{1+\cdots}{2020+\cdots}=\left(\frac{1}{2020}\right)
\end{aligned}

Another approach using Riemann sums.
For integer $a\geq 1$, we have
$$
\frac{1}{n^{a+1}}\sum_{k=1}^n \left(k+a\right)^a
= \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}+\frac{a}{n}\right)^a
\geq \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^a \xrightarrow[n\to\infty]{}\int_0^1 x^a dx = \frac{1}{a+1}
$$
On the other hand, for any fixed $\varepsilon > 0$, we have for $n\geq a/\varepsilon$
$$
\frac{1}{n^{a+1}}\sum_{k=1}^n \left(k+a\right)^a
= \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}+\frac{a}{n}\right)^a
\leq \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}+\varepsilon\right)^a \xrightarrow[n\to\infty]{}\int_0^1 (x+\varepsilon)^a dx = \frac{(1+\varepsilon)^{a+1} - \varepsilon^{a+1}}{a+1}
$$
Therefore, for any fixed $\varepsilon > 0$,
$$
\frac{1}{a+1} 
\leq \lim\inf_{n\to\infty} \frac{1}{n^{a+1}}\sum_{k=1}^n \left(k+a\right)^a
\leq \lim\sup_{n\to\infty} \frac{1}{n^{a+1}}\sum_{k=1}^n \left(k+a\right)^a
\leq \frac{(1+\varepsilon)^{a+1} - \varepsilon^{a+1}}{a+1}
$$
Now take the limit as $\varepsilon \to 0^+$ to conclude.
