Series for $\zeta(3)-\frac{6}{5}$

Question

$\zeta(2)$

The inequality

$$9<\pi^2<10$$

can be obtained from series

$$\zeta(2)=\frac{\pi^2}{6}=\frac{3}{2}+\frac{1}{2}\sum_{k=0}^\infty \frac{1}{(k+1)^2(k+2)^2}\tag{1}$$

and

$$\zeta(2)=\frac{\pi^2}{6}=\frac{5}{3}-\sum_{k=1}^\infty \frac{1}{(k+1)(k+2)^2(k+3)}\tag{2}$$

Both of them have constant numerator and the denominator a polynomial of fourth degree.

$\zeta(3)$

Similarly, for Apéry's constant we can write the inequality

$$\frac{6}{5}<\zeta(3)<\frac{5}{4}$$

from series

$$\zeta(3)=\frac{6}{5}+\sum_{k=2}^\infty \frac{1}{k^3+4k^7}\tag{3}$$

and

$$\zeta(3)=\frac{5}{4}-\sum_{k=0}^\infty \frac{1}{(k+1)(k+2)^3(k+3)}\tag{4}$$

However, the degree of the polynomials is not the same in this case.

Is there a series for $\zeta(3)-\dfrac{6}{5}$ having the denominator a polynomial of order 5?

[EDIT]

From Jack D'Aurizio's study,

$$\begin{align} \zeta(3)&=\frac{26}{21}-\frac{32}{7}\sum_{k=0}^\infty \frac{1}{(2k+1)(2k+3)^3(2k+5)}\\ &=\frac{1138}{945}-\frac{32}{7}\sum_{k=1}^\infty \frac{1}{(2k+1)(2k+3)^3(2k+5)}\\ &=\frac{198862}{165375}-\frac{32}{7}\sum_{k=2}^\infty \frac{1}{(2k+1)(2k+3)^3(2k+5)}\\ &=\frac{9741838}{8103375}-\frac{32}{7}\sum_{k=3}^\infty \frac{1}{(2k+1)(2k+3)^3(2k+5)}\\ &=\frac{2893129886}{2406702375}-\frac{32}{7}\sum_{k=4}^\infty \frac{1}{(2k+1)(2k+3)^3(2k+5)}\\ &=\frac{4550782178678}{3785742835875}-\frac{32}{7}\sum_{k=5}^\infty \frac{1}{(2k+1)(2k+3)^3(2k+5)}\\ \end{align}$$

After James Arathoon's answer,

$$\begin{align} \zeta(3)&=1+\frac{1}{2}\sum_{k=0}^\infty \frac{2k+3}{(k+1)^3(k+2)^3}\\ \\ &=1+\frac{1}{2}\sum_{k=1}^\infty \frac{2k+1}{k^3(k+1)^3}\\ \\ \zeta(3)&=\frac{6}{7}+\frac{64}{7}\sum_{k=0}^\infty \frac{k+1}{(2k+1)^3(2k+3)^3}\\ \\ &=\frac{6}{7}+\frac{64}{7}\sum_{k=1}^\infty \frac{k}{(4k^2-1)^3}\\ \end{align}$$

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$\zeta(3)\approx \frac{6}{5}$ is a consequence of the continued fraction in formula (35), Apéry's Constant from Wolfram MathWorld.

Answer 1

I bet this can be considered as "cheating", but for any $a>0$ we have $$ \sum_{n\geq 0}\frac{1}{(n+a)(n+a+1)^3(n+a+2)} = \frac{1+a}{2a^3}+\frac{1}{2a^3(a+1)}+\frac{1}{2}\psi''(a)\tag{1} $$ by partial fraction decomposition, and $\psi''(1)=\zeta(3), \psi''\left(\frac{1}{2}\right)=-14\,\zeta(3)$, $\psi''\left(\frac{3}{2}\right)=16-14\,\zeta(3), \psi''\left(\frac{5}{2}\right)=16+\frac{16}{27}-14\,\zeta(3)$ and

$$ \psi''\left(m+\frac{1}{2}\right)=16\left[\sum_{k=1}^{m}\frac{1}{(2k-1)^3}-\frac{7}{8}\zeta(3)\right]\tag{2} $$ By choosing a rather large value of $a$ like $a=\frac{7}{2}$, by $(1)$ and $(2)$ we get $$ 7\zeta(3) = \frac{9741838}{1157625}-\sum_{n\geq 0}\frac{1}{\left(n+\frac{7}{2}\right)\left(n+\frac{9}{2}\right)^3\left(n+\frac{11}{2}\right)}\tag{3} $$ where the involved series has a positive value, but less than $10^{-3}$, leading to an accurate approximation for $\zeta(3)$ of the wanted type. Similarly $$2\cdot 10^{-4}\approx\sum_{n\geq 0}\frac{1}{\left(n+\frac{11}{2}\right)\left(n+\frac{13}{2}\right)^3\left(n+\frac{15}{2}\right)}=\frac{4550782178678}{540820405125}-7\zeta(3)\tag{4}$$ and so on. By considering the continued fraction of $\frac{4550782178678}{3785742835875}$ and truncating it at the fourth convergent, we get $\zeta(3)\approx\color{red}{\frac{113}{94}}$ with an approximation error that is less than $10^{-4}$. $\frac{6}{5}$ is just the approximation we get by stopping at the third convergent.

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It is interesting to point out that $\zeta(3)\approx\frac{6}{5}$ arises from the computation of the average order of the arithmetic function $\sigma(n)^2$. Since by Euler's product $$ \forall s>3,\qquad \sum_{n\geq 1}\frac{\sigma(n)^2}{n^s}=\frac{\zeta(s)\zeta(s-1)^2\zeta(s-2)}{\zeta(2s-2)} $$ by tauberian theorems we have $\sum_{n\leq x}\sigma(n)^2 = \frac{5\zeta(3)}{6}x^3+O(x^2)$. On the other hand $3\zeta(4)\approx\zeta(3)\zeta(2)^2$ leads to $\zeta(3)\approx\frac{6}{5}$ and $\sum_{n\leq x}\sigma(n)^2\approx n^3$.

Answer 2

Hint: I don't have enough reputation points to comment on your question which is what I would have preferred to do. I must point out from the information you have provided I do not fully understand the significance of your hard requirement for $\frac{6}{5}$ as a lower bound, so my answer will be minimal and stated without proof at this stage, just in case I am wrong in my understanding of what you are trying to achieve.

Multiplying your $\zeta(3)$ inequality by $20$ gives inequality bounds differing by $1$

$$24<20\zeta(3)<25$$

This is because the difference in the two actual rational bounds used results is an Egyptian Fraction $$\frac{5}{4}-\frac{6}{5}=\frac{1}{20}$$

Is this an essential requirement for an acceptable answer?

Assuming the answer to this question to be yes, my answer is (stated as a hint and without proof)

$$\zeta(3)=\frac{19}{16}+\frac{1}{2}\sum_{k=0}^\infty \frac{2k+5}{(k+2)^3(k+3)^3}$$

The difference between the bounds of the inequality being $\frac{5}{4}-\frac{19}{16}=\frac{1}{16}$; being wider bounds than your original example.

For closer bounds (with the difference also resulting in an Egyptian fraction) you can just increment up both series, with the first step as shown below; which if I understand it was your original intention of asking for a denominator with polynomial order 5.

$\zeta(3)=\frac{29}{24}-\sum_{k=1}^\infty \frac{1}{(k+1)(k+2)^3(k+3)}$ and $\zeta(3)=\frac{259}{216}+\frac{1}{2}\sum_{k=1}^\infty \frac{2k+5}{(k+2)^3(k+3)^3}$

Thus the next difference between bounds is $\frac{29}{24}-\frac{259}{216}=\frac{1}{108}$; being narrower bounds than your original example, with each iteration after that providing closer and closer rational bounds to $\zeta(3)$.