Question: If $s \in \mathbb{N}$ is it true: $\sum_{x=1}^\infty\left[\prod_{k=0}^s{s\choose k}x^k\right]^{-1}={\zeta\left[{s+1 \choose 2}\right] \above 1.5pt \prod_{k=1}^s{s \choose k}};$ where $\zeta\left[{s+1 \choose 2}\right]$ is the Riemann zeta function$^1$ evaluated at the triangular numbers$^2$ ?

I believe I can show the answer to the question is yes by way of inspection. I cannot get an actual explicit calculation that answers the question affirmatively - that is what I am asking for !

Solution to question by inspection: For short handedness I write $$\zeta_{\times}(s)=\sum_{x=1}^\infty\Bigg[\prod_{k=0}^s{s\choose k}x^k\Bigg]^{-1}$$ I computed a small list of values of $\zeta_{\times}(s)$ for small $s$:

\begin{array}{ l | l } s & 2 & 3 & 4 & 5 & 6 & 7\\ \hline \zeta_{\times}(s) & \frac{\zeta(3)}{2} & \frac{\zeta(6)}{9} & \frac{\zeta(10)}{96} & \frac{\zeta(15)}{2500} & \frac{\zeta(21)}{162000}& \frac{\zeta(28)}{26471025}\\ \end{array}

Inspection suggests that $\zeta_{\times}(s)$ has numerator equal to the Riemann zeta function at ${s+1 \choose 2}.$ The denominators of $\zeta_{\times}(s)$ can be written explicitly as $2,9,96,2500,162000,\ldots$; which appear to be the integer sequence A001142$^3$. Listed in the reference section is the following paper by Lagarias: Products of Binomial Coefficients and Unreduced Farey Fractions.$^4$. Lagarias defines the unreduced Farey fractions$^5$ to be the ordered sequence of all reduced and unreduced fractions between $0$ and $1$ with denominator of size at most $s.$ Following Lagarias' notation: I write $G_s$ for the set of unreduced Farey fractions and let $|G_s|$ and $G_s^*$ denote the cardinality and the product of all elements of $G_s$ respectively. Lagarias shows the following results: $|G_s|={s+1 \choose 2}$ and $G_s^*=\prod_{k=1}^s{s \choose k}.$ I put everything together and get $$\zeta_{\times}(s)={\zeta\big(|G_s|\big) \above 1.5pt G_s^*}$$

and I am done. $\blacksquare$

Source and motivation of the problem: Curiosity is the main driving source of the problem. Here it goes: If $x$ and $s$ are positive numbers I write $\zeta(s)$ for the Riemann zeta function evaluated at the number $s.$ For ease of typing I write Pascal's Triangle$^{6}$ like this: $$\text{ }\begin{matrix} 1&&&&\\ 1&1&&&\\ \color{red}{1}&\color{red}{2}&\color{red}{1}&&\\ \color{blue}{1}&\color{blue}{3}&\color{blue}{3}&\color{blue}{1}&\\ 1&4&6&4&1\\ \vdots \end{matrix}$$ and recall that the first row in Pascal's triangle is numbered at $0$. Now look at the second and third rows of the triangle, highlighted in red and blue respectively and observe: $\sum_{x=0}^\infty{ 1 \above 1.5pt \color{red}{1}+\color{red}{2}x+\color{red}{1}x^2}=\zeta(2)$ and $\sum_{x=0}^\infty{ 1 \above 1.5pt \color{blue}{1}+\color{blue}{3}x+\color{blue}{3}x^2+\color{blue}{1}x^3}=\zeta(3).$ This observation shows that $\zeta(s)$ can be recovered from the rows of Pascal's triangle - in particular I can write $\sum_{x=0}^\infty\frac{1}{(1+x)^s}=\zeta(s).$ On the other hand using the binomial theorem$^{7}$ allows me to compute $(1+n)^s$ explicitly with the formula: $\sum_{k=0}^s{s\choose k}x^k$ in which case after substitution: $$\zeta(s)=\sum_{x=0}^\infty\bigg[\sum_{k=0}^s{s\choose k}x^k\bigg]^{-1}$$ Now out of total curiosity I decided to swap the inner summation inside the big brackets and swap it out with a product, noting carefully that if I do that I need to shift the starting point in the outer summation to $1$ otherwise I would be dividing by zero. For short handedness I wrote $$\zeta_{\times}(s)=\sum_{x=1}^\infty\Bigg[\prod_{k=0}^s{s\choose k}x^k\Bigg]^{-1}$$ Numerical inspection suggested $$\zeta_{\times}(s) ={\zeta\left[{s+1 \choose 2}\right] \above 1.5pt \prod_{k=1}^s{s \choose k}}$$ I double checked my numerical hunch against Sloan's Database and encountered the paper by Lagarias. In Lieu of the fact that Lagarias has an explicit formulae for $\log G_s^*$ (see Lagarias paper or reference above for notation) I primarily became interested in computing $\log \zeta_{\times}(s).$ If I could answer the question affirmatively then I would know that $$\log \zeta_{\times}(s)=\log \zeta(|G_s|)-\log(G_s^*)$$

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    $\begingroup$ What is the source of this problem? It's quite an interesting sum. $\endgroup$ Oct 12, 2018 at 17:14

1 Answer 1


$\zeta_{\times(s)} =\sum_{x=1}^\infty\Bigg[\prod_{k=0}^s{s\choose k}x^k\Bigg]^{-1} $

Maybe I'm just being naive, but since $\prod_{k=0}^s{s\choose k}x^k =\prod_{k=0}^s{s\choose k}\prod_{k=0}^sx^k =G_s^*x^{s(s+1)/2} $, $\zeta_{\times}(s) =\sum_{x=1}^\infty \dfrac1{G_s^*}x^{s(s+1)/2} =\dfrac1{G_s^*}\sum_{x=1}^\infty x^{s(s+1)/2} =\dfrac{\zeta(s(s+1)/2)}{G_s^*} $.

Is there anything more to it than this?

  • $\begingroup$ That looks correct. I never thought to split the product ! $\endgroup$
    – Anthony
    Oct 12, 2018 at 17:52

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