Proving an “amazing” claim regarding $\zeta( 3)$ and Apéry's proof

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.

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It's not true when each $a_n$ is $0$ (which currently is an acceptable choice based on how your post is worded). –  KCd Apr 12 '12 at 19:31
It is true when all $a_n = 0$: the $k=1$ term is $1/(x+a_1)$. –  Robert Israel Apr 12 '12 at 20:32
The formula for $\zeta(3)$ is the third listed in Wikipedia under "Many additional series representations have been found, including:" –  Henry Apr 12 '12 at 21:30
@Robert: I agree with you. I wonder if the numerator was $a_1a_2\cdots a_k$ when I read it. If it wasn't, then I just misread it. –  KCd Apr 13 '12 at 1:26
@KCd I edited. My bad. –  Pedro Tamaroff Apr 13 '12 at 1:26
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2 Answers

For every $n\geqslant1$ and every $(x,a_1,\ldots,a_n)$ such that $x\ne -a_k$ for every $k$, $${\sum_{k=1}^n\frac{a_1a_2\cdots a_{k-1}x}{(x+a_1)\cdots(x+a_k)}=1-\frac{a_1a_2\cdots a_{n}}{(x+a_1)\cdots(x+a_n)}}$$ 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 $${\sum_{k}\frac1{a_k}}\ \text{diverges}.$$ 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 has probability $p_k=\frac{x}{x+a_k}$ to succeed. 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 $1-p_k=\frac{a_k}{x+a_k}$. QED.

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And note that if $a_k > 0$ and $x > 0$, $\prod_{k=1}^\infty \frac{a_k}{x+a_k} = 0$ iff $\sum_{k=1}^\infty \frac{x}{x+a_k} = \infty$ –  Robert Israel Apr 12 '12 at 20:28
@Didier What you're saying is that given an inftinite amount of time and experiments, at least 1 must fail, or succeed? –  Pedro Tamaroff Apr 12 '12 at 22:12
What the probabilistic argument really does is to interpret the LHS of your identity as $1/x$ times the probability that at least one experiment succeeds, ever. Thus the identity holds if and only if the probability of zero success is zero. –  Did Apr 13 '12 at 9:47
@Didier That is to say: at least one expermiment must succed? –  Pedro Tamaroff Apr 13 '12 at 17:29
@PeterTamaroff: I strongly disagree with the fact that you modified my answer without signaling it to me. Please do not do this anymore. If you want something to be modified in one of my posts, write a comment and I will deal with the matter. –  Did Jun 8 '12 at 19:46
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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 $n$].

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Some "amazing" things aren't so amazing after all :) –  nbubis Apr 13 '12 at 1:09
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