Mathematical Explanation of Mathematica Summation ${\sum_{n=1}^{\infty}\frac{(2n-1)!}{(2n+2)!}\zeta(2n)}$

Question

From a mathematical point of view, what phenomena that most likely Mathematica Wolfram encountered when calculating: $$ \sum_{n=1}^{\infty}\frac{(2n-1)!}{(2n+2)!}\zeta(2n)\,=\,\color{red}{\frac{2\log(2\pi)-3}{8}+\frac{\zeta(3)}{8\pi^2}} $$ which is incorrect.

While calculating the sum from this question, I noticed that Wolfram result is containing ${\small\,\frac{\zeta(3)}{8\pi^2}\,}$, which is incorrect. Although I realized that this could be a bug, I started to wonder if there are any logical explanation behind this miscalculation! Has Wolfram algorithm encountered something similar to Riemann Rearrangement Theorem?

Doing more investigations, it turns-out that Wolfram is incorrectly miscalculating the closed form of an entire class of zeta summation, except the last case which is correct. $$ \small \begin{align} \sum_{n=1}^{\infty}\frac{\zeta(\alpha\,n)}{(n+a)(n+b)\dots} &= \sum_{n=1}^{\infty}\left[A\frac{\zeta(\alpha\,n)}{n+a}+B\frac{\zeta(\alpha\,n)}{n+b}+\dots\right] = \\ C+\,\sum_{n=1}^{\infty}\frac{\zeta(\alpha\,n)-1}{(n+a)(n+b)\dots} &= \color{darkgreen}{\sum_{n=1}^{\infty}\left[A\frac{\zeta(\alpha\,n)-1}{n+a}+B\frac{\zeta(\alpha\,n)-1}{n+b}+\dots\right]\,+C} \end{align} $$ And with the appearance of this case (the last correct closed form), I believe there is a mathematical explanation regarding a correct summation method or algorithm that gives a kind of systematic incorrect closed form if it applied in a certain way. Appreciating if someone can explore this and alert us regardless of any bug that may exist in any math app. Thanks.

Answer 1

Only a note.

WolframAlpha can calculate Sum[Zeta[2n]*(2n-1)!/(2n+1)!, {n,1,Infinity}]

$\displaystyle \left(=\sum\limits_{n=1}^\infty \frac{\zeta(2n)(2n-1)!}{(2n+1)!}\right)\enspace$ exactly

but can only approximate Sum[Zeta[2n]*(2n)!/(2n+2)!, {n,1,Infinity}]

$\displaystyle \left(=\sum\limits_{n=1}^\infty \frac{\zeta(2n)(2n)!}{(2n+2)!}\right)$.

This is indeed strange, because the difficulty is about the same. WolframAlpha has problems with slow converging series.

Maybe the term $\,\,\text{$\zeta(3)/(8\pi^2$)}\,\,$ is by chance, because the calculation inaccuracy seems to be very close to this term at some point of the calculations.

It's interesting, that WolframAlpha calculates

Sum[Zeta[2n](2n-1)!/(2n+3)!, {n,1,Infinity}]$~\displaystyle \left(=\sum\limits_{n=1}^\infty \frac{\zeta(2n)(2n-1)!}{(2n+3)!}\right)$

exactly including $\,\text{$\zeta(3)/(8\pi^2$)}$, here $\,\text{$9\zeta(3)/(72\pi^2$)}\,$ .

Answer 2

a proof of the correct answer seems appropriate here (I have filed a bug report, since it persists after 7 years)

following a similarish path to in this answer (in which $\sum_{n=2}^\infty\frac{\zeta(n)}{n(n+1)}=\frac{\log\tau-\gamma}2$ arises as a step in showing the equivalence of Raabe's integral and Stirling's approximation), we have (as a special case of the expansion of $\log(x!)$ in polygamma functions) $$\log x!=\sum_{k=1}^\infty\frac{\zeta(k)(-x)^k}k$$ (where we consider $\zeta(1)=\gamma$ by cancelling out the simple pole)

by the generating function DFT trick, $\frac{\log x!+\log(-x)!}2=\sum_{k=1}^\infty\frac{\zeta(2k)(-x)^{2k}}{2k}$, from whence integrating twice and evaluating at $x=1$ produces your integral. Then we use

• $\int_{0\le t\le u\le1}f(t)=\int_{0\le t\le1}(1-t)f(t)$ (iterating this produces Cauchy's iterated integration formula)
• $x!(-x)!=\frac{\pi x}{\sin(\pi x)}$ by Euler's reflection formula

so$$\begin{aligned} &\sum_{k=1}^\infty\frac{\zeta(2k)}{2k(2k+1)(2k+2)}\\ =&\frac12\int_0^1(1-t)\log\left(\frac{\pi t}{\sin(\pi t)}\right)\\ =&\frac12\left(\begin{array} &&\int_0^1(1-t)\log\pi&=\frac{\log\pi}2\\ +&\int_0^1(1-t)\log t&=-\int_0^1t\log\frac1{1-t}=-\sum_{i=1}^\infty\frac1{i(i+2)}=-\frac34\\ -&\int_0^1(1-t)\log\sin(\pi t)\end{array}\right) \end{aligned}$$$\int_0^1(1-t)\log\sin(\pi t)$ can be added to itself backwards to get $\int_0^1\log\sin(\pi t)=-\log2$, so is itself $-\frac{\log2}2$; the entire sum is thus $\frac{2\log\tau-3}8$.

Answer 3

WolframAlpha has its quirks, especially with the zeta function at odd positions. When infinite sums are nested within each other, the results should always be checked with caution. When estimates are used instead of formulas, deviations occur.

Let $m,v\in\mathbb{N}_0$ and $0\leq x < 1$ with $x \neq 0$ for $m\cdot v=0$ .

$$Z_v(x)|_{v\in\mathbb{N}_0}:=\sum\limits_{k=1}^{\infty}\frac{e^{i2\pi xk}}{k^{v+1}}$$

$$F_m(x)|_{m\in\mathbb{N}_0}:=2\sum\limits_{v=1}^{\infty}\frac{x^{2v}\zeta{(2v)}}{2v+m}$$

with

$Z_v(1)= \zeta{(v+1)}$ and $Z_1(1)=\zeta{(2)}=\frac{\pi^2}{6}$

I'm sorry, I don't remember the literature in question. But the following applies:

$$F_0(x)=-\ln{(1-e^{i2\pi x})} + i\pi(x-\frac{1}{2}) + \ln{(2\pi x)}$$

$$F_m(x)|_{m\in\mathbb{N}}=-\ln{(1-e^{i2\pi x})} + \frac{i\pi}{m+1} + \frac{1}{m} - \frac{m!\zeta(m+1)}{(-i2\pi x)^m} + \sum\limits_{v=1}^\infty\binom{m}{v}\frac{v!Z_v(x)}{(-i2\pi x)^v}$$

Be $(a;b;c):=(\frac{1}{2};-1;\frac{1}{2})$ , so that we have $\frac{1}{(2v)(2v+1)(2v+2)}=\frac{a}{2v} + \frac{b}{2v+1} + \frac{c}{2v+2}$ .

With $x\uparrow 1$ we get:

$$\sum\limits_{v=1}^\infty\frac{\zeta(2v)}{(2v)(2v+1)(2v+2)}= \frac{1}{2}(aF_0(1)+bF_1(1)+cF_2(1)) = \frac{\ln(2\pi)}{4} - \frac{3}{8}$$