<s>[INST] A proof that $\frac{\log^k(1)+\log^k(2)+\dotsb +\log^k(n)}{1^k+2^k+\dotsb +n^k} \to 0$ Let $\left(a_n\right)$ be the following sequence: 
$$a_n =\frac{\log^k\left(1\right)+\log^k\left(2\right)+\dotsb +\log^k\left(n\right)}{1^k+2^k+\dotsb +n^k},$$ for a fixed $k \in \mathbb{N}$. Prove that $a_n \to 0$. There are many proofs for this one, but I think this is an elegant one. I'd like you to check it out, first because I consider it to be a nice and concise and elementary, and secondly because I'd like to make sure that there are no mistakes.
Let's get started: First of all, $a_n\geq 0 \ \forall \,n \in \mathbb{N}$. We write $a_n$ as 
$$\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+2^k\left(\frac{\log\left(2\right)}{2}\right)^k+\dotsb +n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}.$$ 
It is obvious that $\left(\frac{\log\left(n\right)}{n}\right)^k \to 0$.  Let $\varepsilon >0$. Then $\exists\, n_0 \in \mathbb{N}$ such that $\forall \, n \geq n_0, \left(\frac{\log\left(n\right)}{n}\right)^k < \varepsilon$. 
We have 
\begin{align*}
a_n&=\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k+ n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb+n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}\\
&=\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+2^k+\dotsb +n^k} + \frac{n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb+n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}\\
&\leq\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+\dotsb +n^k}\\ & \qquad \qquad+ \frac{\varepsilon\left(1^k+\dotsb +\left(n_0-1\right)^k\right)+n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb +n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+\dotsb +n^k}\\
&\leq\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+\dotsb +n^k} + \frac{\varepsilon\left(1^k+2^k+\dotsb +n^k\right)}{1^k+2^k+\dotsb +n^k}\\
&=\frac{1^k+2^k+\dotsb +\left(n_0-1\right)^k}{1^k+2^k+\dotsb +n^k}+\varepsilon.
\end{align*}
Now, by taking the limsup and the liminf as $n \to\infty$, and since 
$$\frac{1^k+2^k+\dotsb +\left(n_0-1\right)^k}{1^k+2^k+\dotsb +n^k}\to 0,$$ we  have 
$$0 \leq \limsup\left(a_n\right) \leq \varepsilon \quad\text{and}\quad 0\leq \liminf\left(a_n\right) \leq \varepsilon.$$ But $\varepsilon$ was arbitrarily small, so 
$$\liminf\left(a_n\right)=\limsup\left(a_n\right)=0=\lim\left(a_n\right).$$
This is more of a discussion and not so much of a question :)
 [/INST] Fix $k\in \mathbb N.$ Let $a_n$ be the numerator, $b_n$ the denominator. Then $b_n \to \infty.$ Time to think about Stolz-Cesaro, which suggests we consider 
$$\tag 1 \frac{a_{n+1}- a_n}{b_{n+1}- b_n} = \frac{\ln^k (n+1)}{(n+1)^k} = \left (\frac{\ln (n+1)}{n+1} \right )^k.$$
Since $[\ln (n+1)]/(n+1) \to 0,$ the right side of $(1)$  $\to 0.$ By Stolz-Cesaro, the limit of interest is $0.$
</s>