It's ok to find the limit with integral
$$\mathop {\lim }\limits_{n \to \infty } \frac{{{1^{2022}} + {2^{2022}} + ... + {n^{2022}}}}{{{n^{2023}}}} = \frac{1}{{2023}}$$ Put ${u_n} = \frac{{{1^{2022}} + {2^{2022}} + ... + {n^{2022}}}}{{{n^{2023}}}} - \frac{1}{{2023}}$
Because $\sum\limits_{i = 1} {\frac{i}{{{n^{2023}}}}} $ is decrease when n $ \nearrow $. Then by using Stolz-Cesaro theorem.
Then
$$\mathop {\lim }\limits_{n \to \infty } n\left( {\frac{{{1^{2022}} + {2^{2022}} + ... + {n^{2022}}}}{{{n^{2023}}}} - \frac{1}{{2023}}} \right)=\lim n{u_n} = \lim \left( {{u_{n + 1}} - {u_n}} \right) = 0$$
But, the key of title is $\dfrac{1}{2}$ not $0$. Can someone help me, thanks
