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 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. 

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 $n$]. 
