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

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This question/ series of questions arose from just curiosity of alternating series.

The Dirichlet eta function is usually defined for all complex input (in the branch of maths known as analytic number theory).

For this question, define the real-valued Dirichlet eta function for $x \in (0, \infty):$ $$\eta\ (x) = \sum_{n=1}^{\infty}\frac{(-1)^{(n+1)}}{n^x} = \frac{1}{1^x} - \frac{1}{2^x} + \frac{1}{3^x} - \frac{1}{4^x} + ... $$

$\eta(1)$ is known to be $\ln(2)$, and this thread concludes that $\eta(\frac{1}{2}) \approx 0.6048986434$

enter image description here

In fact, it appears that the concavity of the Dirichlet eta function was only proved as recently as 2015:

https://www.sciencedirect.com/science/article/pii/S0022314X15000220

I'm interested in whether there's a relatively nice and simple proof that $\eta\ (x)$ is increasing on $x \in (0, \infty)$

My other query is that, surely $\eta\ (x)$ is undefined at $ x = 0 $ because $1 - 1 + 1 - 1 + ...$ does not converge. However, several sources state that $ \eta\ (0) = 1/2$. For example, at the bottom of the Wolfram/Mathworld page on it:

enter image description here

Who is wrong, me or them? (Probably me, but why?)

And finally, assuming I'm right on this, is there a relatively nice and simple proof that $ \lim_{ x \to 0^+ } \eta\ (x) = 1/2$ ?

Progress on first question:

$\eta\ (x) = \frac{1}{1^x} - \frac{1}{2^x} + \frac{1}{3^x} - \frac{1}{4^x} + ... = 1 - 2^{-x} + 3^{-x} - 4^{-x} + ....$

This is increasing $ \impliedby \eta'\ (x) > 0$, so if we can show $\eta'\ (x) > 0 $ for $x \in (0, \infty)$, then we are done.

$\eta'\ (x) = \frac{d}{dx}(1 - 2^{-x} + 3^{-x} - 4^{-x} + ....) = 2^{-x}ln(2) - 3^{-x}ln(3) + 4^{-x}ln(4) + .... = \frac{ln(2)}{2^x} - \frac{ln(3)}{3^x} + \frac{ln(4)}{4^x} - \frac{ln(5)}{5^x} ...$ How do I show this converges to a positive number for all $x>0$?

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Answer to the first part: "Surely $\eta (x)$ is undefined at $x=0$ because $1−1+1−1+... $ does not converge."

By usual way we do series, the sequence of partial sums, the sequence does not converge. However, there are other ways to carry out limits. Wikipedia states that:

The Cesàro sum is defined as the limit, as n tends to infinity, of the sequence of arithmetic means of the first n partial sums of the series.

The Abel summation seems to be perhaps more relevant, but I don't understand it at this time.

See:

en.wikipedia.org/wiki/Dirichlet_eta_function#Particular_values and en.wikipedia.org/wiki/Divergent_series#Abel_summation and en.wikipedia.org/wiki/Ces%C3%A0ro_summation

$$$$

Answer to the second part: See:

https://mathoverflow.net/questions/180716/prove-that-the-dirichlet-eta-function-is-monotonic

However, if someone finds a simpler proof, continuing with my differentiation attempt for example, then of course by all means feel free to post an answer. Edit: in fact, I've asked the derivative method as a separate question here.