Convergence conditions for $\sum\limits_{n\ge1}(-1)^{n-1}\frac{(\log n)^p}n$ where $p\in\mathbb R$

Question

As far as I know, the fact that the series starts at $n=1$ is intentional, so I know I can immediately omit the case where $p\le0$.

I decided to use the alternating series test, so to show that the summand is decreasing in absolute value, I consider the function $f(x)=\dfrac{(\log x)^p}x$ and found that its derivative is positive for $x\in(1,e^p)$ and decreasing for $x>e^p$.

I also find that

$$\lim_{n\to\infty}\frac{(\log n)^p}n=0$$

via L'Hopital's rule.

But barring that doubt, it would seem the series converges for all $p>0$.

Yet when I check via Mathematica, I get

In[1]:= SumConvergence[(-1)^(n-1)Log[n]^p/n, n]

Out[1]= 1 + p < 0

I don't understand this result at all. When $p=0$ the series is clearly convergent and has a value of $\log2$. Larger values of $p$ also return summations in terms of the Euler-Mascheroni and Stieltjes constants. Perhaps this is a bug in the software?

So my question is,

Is my answer correct, and if not, where do I go wrong?

Answer 1

You're on the right track. The limit

$$\lim_{n\to\infty}\frac{(\log n)^p}n=0$$

should be one that's under your belt. How could the measly $\log n,$ no matter if raised to the zillionth power, compete with any power of $n?$ I'm not sure why you're having trouble with L'Hopital. Both top and bottom $\to \infty$ right? So apply monsieur L'Hopital. You get $p(\log n)^{p-1}/n.$ If $p\le 1,$ you're done, as the last expression $\to 0.$ If not, apply LHR again. Clearly after a finite number of phone calls to the good marquis, you'll get a limit of $0.$

Answer 2

There is a simple argument if you are willing to use the integral form of the logarithm. If $\alpha > 0$ then $$\log n = \int_1^n t^{-1} \, dt < \int_1^n t^{-1 + \alpha} \, dt = \frac{n^\alpha - 1}{\alpha} < \frac{n^\alpha}{\alpha}.$$ In particular if $p > 0$ you can take $\alpha = \frac 1{2p}$ to get $$(\log n)^p < (2p)^p n^{\frac 12} $$ and thus $$0 \le \frac{(\log n)^p}{n} < \frac{(2p)^p}{n^{\frac 12}} \to 0$$ as $n \to \infty$.