# Efficient (divergent) summation for sum of zetas at negative arguments?

In a question in MSE (see bottom of my own answer) I'm considering the following series, depending on a parameter m: $$L(m) = -\zeta(1m)/1 - \zeta(2m)/2 - \zeta(3m)/3 - \ldots$$ where I want to make sense of that sums at negative m.

For the numerical evaluation I use a customized version of the Noerlund-summation ($\mathfrak N$) (with the software Pari/GP), but which converges only poorly and even with 128 terms I get not much more than some 15 correct digits. However, at least for $m=-1$ I get a convincing guess using mathematica (at wolfram alpha) such that $$L(-1) \underset{\mathfrak N}{=} 1- \log( \sqrt{2 \pi}) \qquad \small \text{ // guessed from 15 digits }$$ For fractional m between $0 \gt m \gt -1$ the summation still works in principle but with even less reliable digits precision. Here are some guesses for $B(m) = \exp(L(m))$ $$\small \begin{array} {r|lr} m & B(m) (\text{ using } \mathfrak N ) \\ \hline -1 & 1.0844375514192275466 & \qquad (=\exp(1)/ \sqrt{2 \pi} = 1-A_{-1})\\ -1/2 & 1.2904007518681174634 \\ -1/3 & 1.48044921903 \\ -1/4 & 1.65184851943 \\ \end{array}$$

Unfortunately for $m \lt -1$ my procedures seem to be completely useless.
So I'd like to ask here: Q: Which (efficient) procedure can I use to get meaningful evaluations for $L(m)$ at negative m ?

[update]
To respond to @joro's computation: I did also a Borel-summation. The result for $L(-1)$ was $$L(-1) \sim 0.08106146679532725821967026359438236013860...$$

I proceeded this way. My function to be summed at -1 is $$L(m) = - \sum_{k=1}^\infty \zeta(km)/k \qquad \text{ at } m=-1$$

The Borel-transform is $$\mathfrak B L(-1) = - \sum_{k=0}^\infty \zeta(-1-k)/(1+k) \cdot x^k/k!$$ and we define $$B_0(x) = - 1/x \sum_{k=1}^\infty \zeta(-k) \cdot x^k/k!$$ (where the index is also conveniently adapted)

Then the Borel-sum is computed by the integral $$L(-1) \underset{\mathfrak B}{=}\int_0^\infty \exp(-t) B_0(t) dt$$

Now using the software Pari/GP and the sumalt-procedure for $B_0(x)$ we are still confined to small x, so the integral cannot be evaluated at high values of t . But the $B_0(x)$ can be expressed in a closed form using only the exponential, which I denote here as $B_1(x)$: $$B_1(x)= \left(\frac 12- \frac 1x -{1 \over 1-\exp(x)}\right) \cdot \frac 1x$$ This integral can now be evaluated numerically by Pari/GP with a far better interval: $$L(-1) \underset{\mathfrak B}{\sim}\int_{1e-20}^{1e6} \exp(-t) B_1(t) dt$$ and gives the above value to about 30 digits precision (I don't think it is a rational value).

Unfortunately, I cannot generalize that transformation to closed form for sequences of $\zeta(1m)/1,\zeta(2m)/2 , \zeta(3m)/3, ...$ where a negative $m$ is different from $-1$, say $m=-1/2$ or $m=-3$ ... [/update]

-
For m=-1 all sumation methods of mpmath give the rational value 0.0815957190957190957190957190. – joro Apr 19 '13 at 7:59
......for L(-1) – joro Apr 19 '13 at 8:02
@joro: are you sure that are methods for divergent summation? And if, are they strong enough? For instance, using "sumalt" in Pari/GP is not strong enough although it is in general a very useful procedure for divergent summation. – Gottfried Helms Apr 19 '13 at 8:44
joro: mpmath.nsum does not work well if some terms are zero. Excluding those gives 0.081061466795328 (it does not appear possible to get much more accuracy). – Fredrik Johansson Apr 19 '13 at 11:00
ok, I don't claim the answer is correct, just wrote what mpmath returns. – joro Apr 19 '13 at 11:06

This is a partial answer for $m=-1$, proving that Gottfried Helms' guess is correct.

What we need is the following:

When $$B_1(x)= \left(\frac 12- \frac 1x -{1 \over 1-\exp(x)}\right) \cdot \frac 1x,$$

$$\int_0^\infty \exp(-t) B_1(t) dt {=}1- \log( \sqrt{2 \pi})$$

I have used the same idea for this question in MSE: http://math.stackexchange.com/questions/340718/references-to-integrals-of-the-form-int-01-left-frac1-log-x-frac/342072#342072

The idea is considering the following integral:

$$F(s)=\int_0^\infty \left(\frac 12- \frac 1t -{1 \over 1-\exp(t)}\right) \cdot t^{s-1} e^{-t} dt$$

This is well-defined if $\textrm{Re}(s)>-1$. In particular for $s=0$.

This integral can be treated term by term if we have $s$ with sufficiently large real part (absolute convergence).

The result is $$\frac 12 \Gamma(s) - \Gamma(s-1) + \Gamma(s)(\zeta(s)-1)$$

Since the integral defines analytic function on $\textrm{Re}(s)>-1$, the result of integral should be analytic continuation of the function $\frac 12 \Gamma(s) - \Gamma(s-1) + \Gamma(s)(\zeta(s)-1)$.

Now the result for $s=0$ follows if we take limit $s\rightarrow 0$ in that expression.

-
hmm, if I feed the last mentioned function into Pari/GP I get for $s=\epsilon$ the result of $\epsilon^{−1}$ for epsilon near zero – Gottfried Helms Apr 20 '13 at 11:37
That must be an error, this function does have limit as $s\rightarrow 0$, and it is $1+\zeta'(0)$. – i707107 Apr 20 '13 at 16:28
See this: wolframalpha.com/input/… – i707107 Apr 20 '13 at 16:33
I see; I've made a sign-error, sorry. – Gottfried Helms Apr 20 '13 at 18:19