(Real Analysis) Situations where $\lim_{n \rightarrow \infty} \frac{1}{n^\alpha} \sum_{k=1}^n k^\beta \log(k)$ is finite (This exercise is pulled from a past real analysis qualifying exam.) Let $\alpha, \beta$ be nonnegative real numbers. For precisely what set of pairs $(\alpha, \beta)$ do we have
\begin{align}
\lim_{n \rightarrow \infty} \frac{1}{n^\alpha}\sum_{k=1}^n k^\beta \log(k) < \infty ?
\end{align}
Admittedly, I'm not sure what angle to attack this problem from. One possible way is to look at it from a "growth rates" perspective. Pick $\alpha$ and $\beta$ that cause $n^\alpha$ to grow either at the same rate as or faster than $\sum_{k=1}^n k^\beta \log(k)$. But how could I even begin to try and check what $\alpha$'s and $\beta$'s work here? The numerator $\sum_{k=1}^n k^\beta \log(k)$ is a summation, while the denominator $n^\alpha$ is a product.
Here is another perspective I thought of: If we have
\begin{align}
\lim_{n \rightarrow \infty} \frac{1}{n^\alpha}\sum_{k=1}^n k^\beta \log(k) = M
\end{align}
then $\sum_{k=1}^n k^\beta \log(k)$ tends to $Mn^\alpha$ as $n$ tends to $\infty$. But it is difficult for me to imagine any situation where the function starts to "look" like the function on the right. Assume for instance $\beta = 1$. Then, $$\sum_{k=1}^n k \log(k) = \log\left(\prod_{k=1}^n k^k\right)~.$$ 
One more idea is to think of $$f(k) = \frac{1}{n^\alpha}\sum_{k=1}^n k^\beta \log(k)$$ as a measurable funciton over ($\mathbb{N}$,$2^\mathbb{N}$, $c$). Now our problem is to find $\alpha, \beta$ for which $\int_\mathbb{N} f$ is finite over this measure space.
What should I look for to get this problem done? I believe with a well-worded hint I can get this figured out.
 A: The denominator $n^\alpha$ is increasing and unbounded in $n$, so that the Stolz–Cesàro theorem can be applied:
$$
\lim_{n \to \infty} \frac{1}{n^\alpha}\sum_{k=1}^n k^\beta \log(k)
= \lim_{n \to \infty} \frac{n^\beta \log(n)}{n^\alpha - (n-1)^\alpha}
$$
if the latter limit exists (as a finite value or $\infty$). From the mean-value theorem we get that
$$
n^\alpha - (n-1)^\alpha = \alpha n^{\alpha - 1} (1+o(1))
$$
so that
$$
\lim_{n \to \infty} \frac{n^\beta \log(n)}{n^\alpha - (n-1)^\alpha}
= \frac 1 \alpha \lim_{n \to \infty} n^{\beta - \alpha +1} \log(n)
$$
and that is zero if $\beta - \alpha + 1 < 0$, and $\infty$ otherwise. It follows that
$$
\lim_{n \to \infty} \frac{1}{n^\alpha}\sum_{k=1}^n k^\beta \log(k) = 
\begin{cases}
0 & \text{if } \beta - \alpha + 1 < 0 \, ,\\
\infty & \text{if } \beta - \alpha + 1 \ge  0 \, .
\end{cases}
$$
A: From
$$\frac{\log{2}}{n^\alpha}\sum_{k=2}^n k^\beta  <
\frac{1}{n^\alpha}\sum_{k=1}^n k^\beta \log(k) < 
\frac{\log{n}}{n^\alpha}\sum_{k=1}^n k^\beta$$
or
$$\frac{\log{2}}{n^\alpha}\sum_{k=1}^n k^\beta - \frac{\log{2}}{n^\alpha} <
\frac{1}{n^\alpha}\sum_{k=1}^n k^\beta \log(k) < 
\frac{\log{n}}{n^\alpha}\sum_{k=1}^n k^\beta$$
or
$$\frac{n^{\beta+1}\log{2}}{n^\alpha}\color{red}{\frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^\beta} - \frac{\log{2}}{n^\alpha} <
\frac{1}{n^\alpha}\sum_{k=1}^n k^\beta \log(k) < 
\frac{n^{\beta+1}\log{n}}{n^\alpha}\color{red}{\frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^\beta} \tag{1}$$
Now, from this (i.e. Riemann sums)
$$\color{red}{\frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^\beta} \to 
\int\limits_0^1 x^{\beta}dx=\frac{1}{\beta +1}$$
and this
$$\lim\limits_{n\rightarrow\infty}\frac{\log^k{n}}{n^{\varepsilon}}=0$$
We see that 


*

*For $\alpha>\beta+1$ both LHS and RHS of $(1)$ converge to $0$. Thus the sequence in question converges to $0$ as well.

*For $\alpha<\beta+1$ LHS of $(1)$ goes to $+\infty$. Thus the sequence in question goes to $+\infty$.



The final part is $\alpha=\beta+1$. This time we have
$$\frac{1}{n^{\beta+1}}\sum_{k=1}^n k^\beta \log(k)=
\frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^\beta \left(\log\left(\frac{k}{n}\right)+\log{n}\right)=\\
\color{green}{\frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^\beta \log\left(\frac{k}{n}\right)}+\log{n}\color{red}{\frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^\beta}=...$$
and (left as an exercise)
$$\color{green}{\frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^\beta \log\left(\frac{k}{n}\right)} \to \int\limits_0^1 x^{\beta}\log{x} dx =
-\frac{1}{(\beta+1)^2}$$
as a result
$$...\sim -\frac{1}{(\beta+1)^2} + \log{n}\cdot \frac{1}{\beta+1} \to +\infty$$
A: Note that $x^\beta \log x$ is monotonically increasing for $\beta \in [0,\infty)$, hence
$$\sum_{k=1}^{n-1}k^\beta \log k < \int_1^nx^\beta \log x\;dx < \sum_{k=2}^nk^\beta \log k = \sum_{k=1}^nk^\beta \log k$$
or with obviously defined notation: $$S_{n-1}(\beta) < I_n(\beta) < S_n(\beta)$$
Using integration by parts we find
\begin{align*}
I_n(\beta) &= \int_1^n \underbrace{\log x}_{u}\underbrace{x^\beta\;dx}_{dv} =
\frac{n^{\beta+1}  \log n}{\beta+1}
- \frac{1}{\beta+1} \int_1^nx^{\beta+1}\;\frac{dx}{x}\\
&= \frac{n^{\beta+1}\log n}{\beta+1} - \frac{n^{\beta+1}-1}{(\beta+1)^2}\\
&=\frac{1}{\beta+1}\left(\log n - \frac{1}{\beta+1} \right) n^{\beta+1}
+ \frac{1}{(\beta+1)^2}
\end{align*}
then clearly
$$\lim_{n\to\infty}\frac{I_n(\beta)}{n^\alpha} = \begin{cases}
\infty &\text{if $\beta-\alpha+1 \geq 0$,}\\
0 & \text{if $\beta-\alpha+1 < 0$.}
\end{cases}$$
If $\beta-\alpha+1 \geq 0$ then
$$\frac{I_n(\beta)}{n^\alpha}
< \frac{S_n(\beta)}{n^\alpha} \to \infty$$
whereas if $\beta-\alpha+1 < 0$
$$
0 < \frac{S_n}{n^\alpha}
= \frac{S_{n-1} +  n^\beta \log n}{n^\alpha}
< \frac{I_n}{n^\alpha} + \underbrace{n^{\beta-\alpha+1} \frac{\log n}{n}}_{\to\, 0} \to 0
$$
