# I can Euler-sum $\sqrt{-\ln(1)}-\sqrt{-\ln(2)}+\sqrt{-\ln(3)}-\cdots$. But how can I do $\sqrt{-\ln(1)}+\sqrt{-\ln(2)}+\sqrt{-\ln(3))}+\cdots$?

This is also related to an older thread in MSE ("what is the half derivative of zeta at zero?") .

One of the possible steps in the problem of that thread was to evaluate the series

$$s_a=\eta^{(0.5)}(0) = \small{\sqrt{-\ln(1)}-\sqrt{-\ln(2)}+\sqrt{-\ln(3)}-\sqrt{-\ln(4)} \cdots \underset{\mathfrak E}{\approx}} - 0.347006596200 i$$ as the (regularized by Euler-summation $\mathfrak E$) value for the half-derivative of the alternating zeta (or "Dirichlet's eta").

Q: But how can I do the nonalternating series $$s_p=\zeta^{(0.5)}(0)= \small{ \sqrt{-\ln(1)}+\sqrt{-\ln(2)}+\sqrt{-\ln(3)}+\sqrt{-\ln(4)}+ \cdots } \underset{\mathfrak ???}{\approx} ???$$ $\qquad \qquad$ I don't see so far any possibility for instance in the sense of L. Euler's famous $\eta() \to \zeta()$ - conversion.

[update]: what I've additionally just done was to apply a procedure to find the/a formal power series for the problem of finite sums of consecutive terms. The procedure is that of the approximation of the Neumann-series of the Carleman-matrix for the function $f(x) = \sqrt{\ln(\exp(x^2)+1)}$ which is the iterative transfer function which produces the term of the series for index $n+1$ from the term at index $n$. The top-left of the heuristically approximated matrix $A$ is $$\small A_{0..16,0..1}=\begin{bmatrix} ?? & ?? \\ 0 & 1 \\ -1.00000000000 & -0.662055270527 \\ 0 & 0 \\ -0.500000000000 & -0.561866397242 \\ 0 & 0 \\ -0.166666666667 & -0.249581408503 \\ 0 & 0 \\ -0.0416666666667 & -0.0755503260124 \\ 0 & 0 \\ -0.00833333333333 & -0.0172091887343 \\ 0 & 0 \\ -0.00138888888889 & -0.00315760368955 \\ 0 & 0 \\ -0.000198412698413 & -0.000499047959470 \\ 0 & 0 \\ -0.0000248015873016 & -0.0000687607442729 \\ ... & ... \end{bmatrix}$$ Using the coefficients of the first column for a power series in $x$ building the function $a_0(x)$ we get $$a_0( \sqrt{\ln(n_1)})-a_0( \sqrt{\ln(n_2)}) = n_2 - n_1$$ which indicates the sum of the $f(x)^0$ which is just counting the terms.

Using the coefficients $a_{k,1}$ of the second column for a power series in $x$ building the function $a_1(x)$ we get the finite sum of the function at consecutive arguments: $$a_1( \sqrt{\ln(n_1)})-a_1( \sqrt{\ln(n_2)}) = \sum_{k=1}^\infty (\sqrt{\ln(n_1)}^k - \sqrt{\ln(n_2)}^k )a_{k,1} = \sum_{k=n_1}^{n_2-1} \sqrt{\ln (k)}$$ which indicates the sum of the $f(x)$ which is the desired finite sums $s_p$ for the required terms.
However - I'm missing the first coefficient for the second power series. That should just contain the representative value for the infinite sum from $n_1=1$ to $\infty$

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Very interesting sum, but I can only half understand what is going on. – Simple Art Jun 16 at 15:50
What do you need? What a "half-derivative" is? – Gottfried Helms Jun 16 at 19:41
No, I understand what fractional derivatives are, but familiar with anything starting from the UPDATE section, which is half the post. – Simple Art Jun 17 at 11:42
I see... Well, this is a method which needs some more background I'm afraid. The idea is, to derive a solution by converting the problem of a powerseries into one containing iterates of a function, which I've applied several times to similar problems. A related link is go.helms-net.de/math/divers/BernoulliForLogSums.pdf where I try to explain what I'm also doing here. Perhaps a more basic introduction is this which deals with the sums-of-like-powers-problem with this method: go.helms-net.de/math/binomial_new/04_3_SummingOfLikePowers.pdf , the $\eta()$-function. – Gottfried Helms Jun 17 at 12:23
@simpleArt : This is not a complete "private" method, one can find application of this under the term "indefinite summation" and there are question&answers here in MSE as well as in MO which cover this subject. Unfortunately I've no good idea at the moment for a good historical text, maybe the Euler-Maclaurin-formula is a usable example, but I've nothing at the top of my head in the moment... – Gottfried Helms Jun 17 at 12:25