Proof that $\sum\limits_{k=1}^\infty\frac{a_1a_2\cdots a_{k-1}}{(x+a_1)\cdots(x+a_k)}=\frac{1}{x}$ regarding $\zeta(3)$ and Apéry's proof

Question

I recently printed a paper that asks to prove the "amazing" claim that for all $a_1,a_2,\dots$

$$\sum_{k=1}^\infty\frac{a_1a_2\cdots a_{k-1}}{(x+a_1)\cdots(x+a_k)}=\frac{1}{x}$$

and thus (probably) that

$$\zeta(3)=\frac{5}{2}\sum_{n=1}^\infty {2n\choose n}^{-1}\frac{(-1)^{n-1}}{n^3}$$

Since the paper gives no information on $a_n$, should it be possible to prove that the relation holds for any "context-reasonable" $a_1$? For example, letting $a_n=1$ gives

$$\sum_{k=1}^\infty\frac{1}{(x+1)^k}=\frac{1}{x}$$

which is true.

The article is "A Proof that Euler Missed..." An Informal Report - Alfred van der Poorten.

Answer 1

For every $n\geqslant1$ and every $(x,a_1,\ldots,a_n)$ such that $x\ne -a_k$ for every $k$, $$ \color{red}{\sum_{k=1}^n\frac{a_1a_2\cdots a_{k-1}x}{(x+a_1)\cdots(x+a_k)}=1-\frac{a_1a_2\cdots a_{n}}{(x+a_1)\cdots(x+a_n)}} $$ Hence the formula in the post holds if and only if $$ \prod_{k=1}^{\infty}\frac{a_k}{x+a_k}=0, $$ which, for $x\gt0$ and at least if the sequence $(a_k)$ is nonnegative, is equivalent to the fact that $$ \color{green}{\sum_{k}\frac1{a_k}}\ \text{diverges}. $$ Here is a probabilistic proof of the finitary version, valid for every nonnegative $a_k$ and positive $x$ (note that once one knows these two rational expressions in $(x,a_1,\ldots,a_n)$ coincide for these values, one knows they are in fact identical).

Proof: Assume that one performs a sequence of $n$ independent experiments and that the $k$th experiment succeeds with probability $$p_k=\frac{x}{x+a_k}.$$ Then the $k$th term of the sum on the LHS of the equation above is the probability that every experiment from $1$ to $k-1$ failed and that experiment $k$ succeeded. Hence their sum is the probability of the disjoint union of these events, which is exactly the event that at least one experiment from $1$ to $n$ succeeded. The complementary event corresponds to $n$ failures, hence its probability is the product from $1$ to $n$ of the probabilities of failures $1-p_k$, that is, $$\prod_{k=1}^n(1-p_k)=\prod_{k=1}^n\frac{a_k}{x+a_k}.$$ This proves the claim.

Answer 2

Formally, the first identity is repeated application of the rewriting rule

$$\dfrac 1 x = \dfrac 1 {x+a} + \dfrac {a}{x(x+a)} $$

to its own rightmost term, first with $a = a_1$, then $a=a_2$, then $a=a_3, \ldots$

The only convergence condition on the $a_i$'s is that the $n$th term in the infinite sum go to zero. [i.e. that $a_1 a_2 \dots a_n / (x+a_1)(x+a_2) \dots (x+a_n)$ converges to zero for large $n$].