(Real Analysis) Situations where $\lim_{n \rightarrow \infty} \frac{1}{n^\alpha} \sum_{k=1}^n k^\beta \log(k)$ is finite

Question

(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.

Answer 1

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

Answer 2

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 $+\infty$.

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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 $$...\sim -\frac{1}{(\beta+1)^2} + \log{n}\cdot \frac{1}{\beta+1} \to +\infty$$

Answer 3

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 $\beta-\alpha+1 < 0$

$$ 0 < \frac{S_n}{n^\alpha} = \frac{S_{n-1} + n^\beta \log n}{n^\alpha} < \frac{I_n}{n^\alpha} + \underbrace{n^{\beta-\alpha+1} \frac{\log n}{n}}_{\to\, 0} \to 0 $$