The inequality


can be obtained from series

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


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

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


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


from series

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


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

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?


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,

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

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

  • 1
    $\begingroup$ I think the best way for approximating $\zeta(3)$ is to exploit Apery's acceleration formula (proved here - math.stackexchange.com/questions/2331692/… - by elementary means) $$\zeta(3)=\frac{5}{2}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\binom{2n}{n}}$$ where the terms of the involved series behave like $\frac{1}{n^{5/2}4^n}$, leading to linear convergence. $\endgroup$ – Jack D'Aurizio Jun 23 '17 at 11:38
  • $\begingroup$ That is a different type of formula. I was curious whether there is a series with a convergence between those of (3) and the definition $\sum_{k>0} \frac{1}{k^3}$, filling the gap suggested by (1) (2) and (4). $\endgroup$ – Jaume Oliver Lafont Jun 23 '17 at 11:56
  • $\begingroup$ Well, you can always use partial fractions on $(3)$ or $(4)$ to write them as sums with denominators at most cubic. $\endgroup$ – Macavity Jun 23 '17 at 12:14
  • $\begingroup$ @JaumeOliverLafont: please have a look at my updated answer. It involves a polynomial with degree $7\neq 5$ but I guess it gets closer to what you are looking for. $\endgroup$ – Jack D'Aurizio Sep 28 '17 at 16:45
  • $\begingroup$ @JackD'Aurizio Yes, but it is equivalent to formula (3) in the question.... $\endgroup$ – Jaume Oliver Lafont Sep 28 '17 at 21:19

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.

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 $\sum_{n\leq x}\sigma(n)^2\approx n^3$.

  • $\begingroup$ This leads to approximations from above as formula (4) in the question, also with degree $5$, but the wanted type $\frac{6}{5}$ is an approximation from below... $\endgroup$ – Jaume Oliver Lafont Jun 23 '17 at 12:38
  • $\begingroup$ @JaumeOliverLafont: from $(4)$ you get that the distance between $\zeta(3)$ and $\frac{4550782178678}{3785742835875}$ is less than $10^{-4}$, hence you get both an upper and a lower bound, where is the issue? $\endgroup$ – Jack D'Aurizio Jun 23 '17 at 13:10
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    $\begingroup$ @JackD'Aurizio, it seems to me what the OP wants is a formula of the form $\zeta(3)-{6\over5}=\sum{1\over P(k)}$, where $P(k)$ is a polynomial of degree $5$ with rational coefficients which is positive for all $k$ in the sum. Preferably $P$ should be monic with integer coefficients, and ideally its roots should all be integers. $\endgroup$ – Barry Cipra Jun 23 '17 at 13:47
  • $\begingroup$ @BarryCipra: I was not able to achieve that, but I achieved the same thing with $\frac{6}{5}$ being replaced by a better approximation of $\zeta(3)$, so I guess we might be equally happy after all. $\endgroup$ – Jack D'Aurizio Jun 23 '17 at 13:49
  • $\begingroup$ @JackD'Aurizio It is a better approximation, but from above, so your formulas (3) and (4) lead to an expression $$\zeta(3)=\frac{6}{5}+ positive fraction - series$$ while the goal was $$\zeta(3)= \frac{6}{5} (+ positive fraction) + series$$ as in (3) from the question. $\endgroup$ – Jaume Oliver Lafont Jun 23 '17 at 23:06

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 $1$


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 $\zeta(3)$.

  • $\begingroup$ The fraction is $\frac{259}{216}$, isn't it? I had not noticed the unit fraction property, but it is a consequence of numbers $4,5,5,6$ matching $n,n+1,n+2$, therefore $n(n+2)=(n+1)^2+1$, so it seems meaningful. From your series, we may want to write $$\zeta(3)=\frac{6}{5}+\frac{19}{17280}+\frac{1}{2}\sum_{k=2}^\infty \frac{2k+5}{(k+2)^3(k+3)^3}$$ which has an extra fraction between $\frac{6}{5}$ and the series. $\endgroup$ – Jaume Oliver Lafont Jul 2 '17 at 13:49
  • $\begingroup$ Updated to correct fraction to $\frac{259}{216}$ $\endgroup$ – James Arathoon Jul 2 '17 at 16:51
  • $\begingroup$ Another series with $\frac{19}{16}$ is $$\zeta(3)=\frac{19}{16}+\sum_{k=0}^\infty \frac{1}{(k+1)^2(k+2)^3(k+3)^2}$$ $\endgroup$ – Jaume Oliver Lafont Aug 6 '17 at 5:18
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    $\begingroup$ Got even closer to $\frac{6}{5}$, but unfortunately a little on the high side with this even faster converging series that I found using Mathematica $$\zeta(3)=\frac{101}{84}-\frac{1}{84} \sum_{k=1}^\infty \frac{16(k(k+1)+3}{3(2k+1)^3(k(k+1))^4}$$ $\endgroup$ – James Arathoon Mar 13 '18 at 19:53

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