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The Dirichlet eta function $\eta(s)$ is related to the Riemann zeta function $\zeta(s)$ as illustrated in (1) below. References (1) and (2) claim formula (2) for $\zeta(s)$ is globally convergent (except where $s=1+\frac{2\,\pi\,i}{\log(2)}n$ and $n\in\mathbb{Z}$) which seems to imply formula (3) for $\eta(s)$ is globally convergent. This is consistent with an answer to one of my related questions posted at reference (3) which claims formula (3) is valid for all $s$.


(1) $\quad\eta(s)=\left(1-2^{1-s}\right)\zeta(s)$

(2) $\quad\zeta(s)=\frac{1}{1-2^{1-s}}\sum\limits_{n=0}^\infty\frac{1}{2^{n+1}}\sum\limits_{k=0}^n\binom{n}{k}\frac{(-1)^k}{(k+1)^{s}}$

(3) $\quad\eta(s)=\sum\limits_{n=0}^N\frac{1}{2^{n+1}}\sum\limits_{k=0}^n\binom{n}{k}\frac{(-1)^k}{(k+1)^s}\,,\quad N\to\infty$


Reference (1): Wikipedia Article: Riemann zeta function, Representations, Globally convergent series

Reference (2): Sondow, Jonathan and Weisstein, Eric W. "Riemann Zeta Function." From MathWorld--A Wolfram Web Resource.

Reference (3): Answer to Questions on two Formulas for $\zeta(s)$


Figure (1) below illustrates the error in formula (3) for $\eta(s)$ evaluated at $N=400$. Note formula (3) for $\eta(s)$ seems to diverge more and more as $s$ becomes increasingly negative.


Error in Formula (3) Evaluated at N=400

Figure (1): Error in Formula (3) for $\eta(s)$ Evaluated at $N=400$


Figure (2) below illustrates a discrete plot of the error in formula (3) for $\eta(s)$ evaluated at integer values of $s$ and $N=1000$. Note formula (3) for $\eta(s)$ actually seems to converge better at negative integers than at positive integers.


Error in Formula (3) Evaluated at Integer Values of s and N=1000

Figure (2): Error in Formula (3) for $\eta(s)$ where $s\in \mathbb{Z}$ Evaluated at $N=1000$


Figures (3) to (6) below illustrate the error in formula (3) for $\eta(s)$ evaluated at $s=-9.5$ over several ranges of $N$. Note the divergence range of formula (3) for $\eta(s)$ evaluated at $s=-9.5$ seems to increase as the evaluation limit $N$ increases.


Error in formula (3) evaluated at s=-9.5 for 0<=N<=100

Figure (3): Error in formula (3) for $\eta(s)$ evaluated at $s=-9.5$ for $0\le N\le 100$.


Error in formula (3) evaluated at s=-9.5 for 0<=N<=200

Figure (4): Error in formula (3) for $\eta(s)$ evaluated at $s=-9.5$ for $0\le N\le 200$.


Error in formula (3) evaluated at s=-9.5 for 0<=N<=400

Figure (5): Error in formula (3) for $\eta(s)$ evaluated at $s=-9.5$ for $0\le N\le 400$.


Error in formula (3) evaluated at s=-9.5 for 0<=N<=800

Figure (6): Error in formula (3) for $\eta(s)$ evaluated at $s=-9.5$ for $0\le N\le 800$.


Question: What is the explanation for the apparent discrepancy between claimed and observational convergences of formula (3) for the Dirichlet eta function $\eta(s)$?

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  • $\begingroup$ The convergence is not uniform perhaps? $\endgroup$
    – Somos
    Apr 10, 2019 at 20:03
  • $\begingroup$ @Somos The convergence doen't seem to be pointwise either. I don't understand in what sense formula (3) is claimed to be globally convergent. $\endgroup$ Apr 10, 2019 at 20:43
  • $\begingroup$ Okay, so, which specific value or values of $s$ is equation $(3)$ inot convergent? $\endgroup$
    – Somos
    Apr 10, 2019 at 21:06
  • $\begingroup$ @Somos For example, Figures (3) to (6) illustrate the error in formula (3) when evaluated at $s=-9.5$ as a function of the upper evaluation limit $N$. $\endgroup$ Apr 10, 2019 at 21:23

1 Answer 1

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You have to be careful with numerical computations. If you are summing positive and negative values, you may get massive loss of significance. Perhaps an example will show what can happen. I use PARI/GP for the calculations. First, define the $\ \eta(s)\ $ function in terms of $\ \zeta(s)\ $ if $N=0$ and using the double sum in equation $(3)$ if $N>0$.

Eta(s, N=0) = {if( N<1, (1 - 2^(1-s)) * zeta(s), sum(n=0, N,
    2^(-n-1) * sum(k=0, n, binomial(n, k) * (-1)^k/(k+1)^s, 0.)))};

Next, try it with low precision and see how the values differ.

? default(realprecision, 19)
? forstep(n=50, 600, 50, print(n, " ", Eta(-9.5) - Eta(-9.5,n)))
50 -9.642528737400027361E-6
100 0.04774435040966354144
150 2.599876523165513738
200 -2.964487980721362893
250 256.1738173836702262
300 35.26046969887404046
350 -4458.254870234773912
400 -9841.293439755364521
450 75026.15715491652695
500 208518.5008905734908
550 249654.0022175838606
600 -194943.3625446287684

Now, try it again but with double precision and see what happens.

? default(realprecision, 38)
? forstep(n=50, 600, 50, print(n, " ", Eta(-9.5) - Eta(-9.5,n)))
50 1.36634363860781380424739243811E-17
100 -9.168029132151541870E-22
150 1.0314039806014013156E-19
200 -1.2323365675288983452001952305E-18
250 -1.07380641622270909181919052693E-17
300 3.2225691859129092780110367112E-17
350 -6.2902053300577065279589792889E-16
400 -1.1634529955480626497925160353E-16
450 4.6382289819863037395153447751E-15
500 8.4040100485998106924892434233E-15
550 -6.8760525356739577517253084299E-15
600 1.24908747773726136990750433575E-14

Notice the huge errors in low precision are gone in double precision. However, the errors still increase with increasing $N$ for a fixed precision. So what you need to do is increase both the precision and the $N$ to get convergence.

P.S. For proof of convergence see the answer to MSE question 3033238 "Questions on two formulas for $\zeta(s)$" in case you are rightfully wary relying on limited numerical evidence.

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  • $\begingroup$ For the proof I had thisone $\endgroup$
    – reuns
    Apr 11, 2019 at 18:08

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