Prove that the series $\sum\limits_{n=2}^{\infty}(-1)^n\frac{\ln(n)}{n^x}$ converges to a positive real number for all $x > 0$

Question

Prove that the series $\sum\limits_{n=2}^{\infty} (-1)^n\frac{\ln(n)}{n^x}=\frac{\ln(2)}{2^x} - \frac{\ln(3)}{3^x} + \frac{\ln(4)}{4^x} - \frac{\ln(5)}{5^x} + ...$ converges to a positive real number for all $x > 0$.

This question arose from this one:

Relatively simple proof that real-valued Dirichlet eta function is increasing, and also a query of the function at / about $x = 0$

Update:

There is a number $\alpha \in (1,2)$ such that for each $x > \alpha,$ the absolute sequence $(a_n)_{n\geq2} = \left(\frac{\ln(n)}{n^x}\right)_{n\geq2} \ $is strictly decreasing. So for $x > \alpha$, the alternating series converges to a positive number.

$x \leq \alpha, \ $ in particular when $x << 1$ is more difficult.

I investigated the series for different value of $x$ below $0$ and it seems that the obvious approach is to show that $\exists$ an odd number $j$ such that $\frac{\ln(2)}{2^x} - \frac{\ln(3)}{3^x} + \frac{\ln(4)}{4^x} - \frac{\ln(5)}{5^x} + ... + \frac{\ln(j-1)}{(j-1)^x} - \frac{\ln(j)}{j^x} > 0,$ and that the absolute sequence $\left(\frac{\ln(n)}{n^x}\right)_{n\geq j}$ is strictly decreasing. I think that, even for the lowest value of $j$ for a given $x$, the absolute sequence will be decreasing. However, this may need to be proven as well.

$x = \frac{12}{20} \implies j = 51$

You can certainly check with a program, terms for "large enough $x$ " i.e. for $x \in (0.5, \alpha)$, but for very small positive $x$ you cannot check with a program. So we probably have to prove it using by rearranging the first $j$ terms and using inequalities to show that the sum of the first $j$ terms is $>0$. And yes, you also need to prove that the absolute sequence is strictly decreasing after the $j-th$ term also, but I don't think this will be as difficult as proving the sum of the first $j$ terms is $>0$.

I have also done a table showing x and values of $j$ where the sum is first positive.

\begin{array}{|c|c|c|c|} \hline x& j\\ \hline 1&5 \\ \hline \frac{19}{20}&7 \\ \hline \frac{18}{20}&7 \\ \hline \frac{17}{20}&9 \\ \hline \frac{16}{20}&11 \\ \hline \frac{15}{20}&15 \\ \hline \frac{14}{20}&21 \\ \hline \frac{13}{20}&31 \\ \hline \frac{12}{20}&51 \\ \hline \frac{11}{20}&89 \\ \hline \frac{10}{20}&181 \\ \hline \frac{9}{20}&441 \\ \hline \frac{8}{20}&1395 \\ \hline \frac{7}{20}&6483 \\ \hline \frac{6}{20}&54805 \\ \hline \end{array}

I made this table by manually inputting values into WA, so it may in fact be incorrect. A Python / C++ program or something may ensure more accurate results?

The bottom line so far: we need to prove $ \exists j \in \mathbb{N}$ for every $x$ with $0< x <\alpha$.

Answer 1

CW

positivity is immediate when $x>1$

Relation to zeta

alright, there is a functional equation for the derivative,

Answer 2

I doubt there is any formula telling immediately that it is non-negative. What I mean is that sometimes you need numerical checks to know the sign of a given series.

For $s > 0$

$$\sum_{n=1}^\infty (-1)^{n+1} n^{-s}= \sum_{n=1}^\infty (\sum_{m=1}^n (-1)^{m+1} ) (n^{-s}-(n+1)^{-s})=\sum_{n=1}^\infty \frac{1+(-1)^{n+1}}{2} (n^{-s}-(n+1)^{-s})$$ $$=\frac12+\frac{s}2 \sum_{n=1}^\infty (-1)^{n+1} \int_n^{n+1} t^{-s-1}dt$$ The latter converges for $s >-1$ and defines the analytic continuation as well as of the derivative.

The uniform convergence for $s\ge 0$ means that you can easily prove, with numerical checks, that $$\eta'(s)=\frac{1}2 \sum_{n=1}^\infty (-1)^{n+1} \int_n^{n+1} t^{-s-1}dt-\frac{s}2 \sum_{n=1}^\infty (-1)^{n+1} \int_n^{n+1}\log(t) t^{-s-1}dt$$ is non-negative on small enough intervals. Doing so with enough intervals you will cover $[0,T]$. Next for $T$ large enough it is non-negative on $[T,\infty)$ because $\eta'(s)=\log(2)2^{-s}+O(3^{-s})$.