(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.
Best Answer
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 $$ \lim_{n \to \infty} \frac{1}{n^\alpha}\sum_{k=1}^n k^\beta \log(k) = \begin{cases} 0 & \text{if } \beta - \alpha + 1 < 0 \, ,\\ \infty & \text{if } \beta - \alpha + 1 \ge 0 \, . \end{cases} $$