Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\infty$

share|cite|improve this question
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.