# Calculating the $\sum_{x=1}^\infty\left[\prod_{k=0}^s{s\choose k}x^k\right]^{-1}$

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 $$\log \zeta_{\times}(s)=\log \zeta(|G_s|)-\log(G_s^*)$$

• What is the source of this problem? It's quite an interesting sum. – Carl Schildkraut Oct 12 '18 at 17:14

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