Problem
Evaluate $$\lim_{n \to \infty}\frac{1^{\lambda n}+2^{\lambda n}+\cdots+n^{\lambda n}}{n^{\lambda n}}$$ where $\lambda>0.$
Solution
Denote $$S_n:=\sum_{k=1}^{n}\left(\frac{k}{n}\right)^{\lambda n}=\sum_{k=0}^{n-1}\left(1-\frac{k}{n}\right)^{\lambda n}.\tag1$$
On one hand, we choose some $p\in \mathbb{N+}$, fix it, and let $n>p+1$. Then
$$S_n \geq \sum_{k=0}^{p}\left(1-\frac{k}{n}\right)^{\lambda n}.\tag2$$
Let $n \to \infty$ within $(2)$. We obtain
$$\liminf_{n \to \infty}S_n\geq \liminf_{n \to \infty} \sum_{k=0}^{p}\left(1-\frac{k}{n}\right)^{\lambda n}=\sum_{k=0}^{p}\liminf_{n \to \infty}\left(1-\frac{k}{n}\right)^{\lambda n}=\sum_{k=0}^{p}e^{-\lambda k}.\tag3$$
Notice that $(3)$ holds for all $p$. We may let $p \to \infty$ and obtain
$$\liminf_{n \to \infty}S_n\geq \sum_{k=0}^{\infty}e^{-\lambda k}=\frac{e^{\lambda}}{e^{\lambda}-1}.\tag4$$
On the other hand, from $1+x\leq e^x$ we may derive $\left(1-\dfrac{k}{n}\right)^{\lambda n}\leq e^{-\lambda k}.$ Thus $$S_n \leq \sum_{k=0}^{n-1}e^{-\lambda k}.\tag5$$ Let $n \to \infty$ within $(5)$. We obtain $$\limsup_{n \to \infty}S_n \leq \limsup_{n \to \infty}\sum_{k=0}^{n-1}e^{-\lambda k}=\frac{e^{\lambda}}{e^{\lambda}-1}.\tag6$$ See $(4)$ and $(6)$. We may conclude $$\frac{e^{\lambda}}{e^{\lambda}-1}\leq \liminf_{n \to \infty}S_n\leq \limsup_{n \to \infty}S_n\leq \frac{e^{\lambda}}{e^{\lambda}-1},\tag7$$ which implies $$\lim_{n \to \infty}S_n=\frac{e^{\lambda}}{e^{\lambda}-1}.\tag8$$
Please correct me if I'm wrong. Thanks.