Hey guys I was reading Alfred van der Poortens paper regarding Apery's constant and I came across this pretty equality. For all $a_1, a_2, \ldots$
\begin{align} \large\sum_{k=1}^{\infty}\frac{a_1a_2\cdots a_{k-1}}{(x+a_1)\cdots (x+a_k)} = \frac 1x \end{align}
I tried fiddling around with the the summands by trying to use a partial fraction decomposition but it just got messy.
link to paper: http://maths.mq.edu.au/~alf/Humid%20Summer/45.pdf
In the context of the paper, we can assume that $x$ is an integer, and the sequence $a_k$'s is an integer sequence such that: $\displaystyle \lim_{K \to \infty} \dfrac{\displaystyle \prod_{i=1}^K a_i}{x\displaystyle \prod_{i=1}^K (x+a_i)} = 0$, then:
As in the paper, he denoted $A_n = \dfrac{\displaystyle \prod_{i=1}^n a_i}{x\displaystyle \prod_{i=1}^n (x+a_i)}$ for $n \geq 1$, and $A_0 = \dfrac{1}{x}$, then:
$\displaystyle \sum_{k=1}^\infty \dfrac{a_1a_2...a_{k-1}}{x(x+a_1)(x+a_2)...(x+a_k)} = \displaystyle \lim_{N \to \infty} \sum_{k=1}^N \dfrac{a_1a_2...a_{k-1}}{x(x+a_1)(x+a_2)...(x+a_k)} = \displaystyle \lim_{N \to \infty} \sum_{k=1}^N (A_{k-1} - A_k) = \displaystyle \lim_{N \to \infty} (A_0 - A_N) = A_0 - 0 = \dfrac{1}{x}$
No matter how I look at it, this seems false. Given values $a_1, a_2, \ldots, a_i$, we can choose $a_{i+1}$ so that the $(i+1)$th term of the series is as small as we like. In particular, we can choose a sequence of $a_i$s so that the series we're summing is bounded by $1/4, 1/8, 1/16, \ldots$ for $x=1$, so the sum is $\le 1/2$ when it should be $1$.
$$(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... =A(x)\tag 1$$
$$\frac{A(x)}{A(x)}=1=x\frac{(x+a_2)(x+a_3)(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... }+a_1\frac{(x+a_2)(x+a_3)(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... } \tag 2$$
$$1=\frac{x}{(x+a_1)}+a_1\frac{(x+a_2)(x+a_3)(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... } \tag 3$$
$$1=\frac{x}{(x+a_1)}+a_1\frac{x(x+a_3)(x+a_4)(x+a_5)...+a_2(x+a_3)(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... } \tag 4$$
$$1=\frac{x}{(x+a_1)}+\frac{a_1x}{(x+a_1)(x+a_2)}+a_1a_2\frac{(x+a_3)(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... } \tag 5$$
$$1=\frac{x}{(x+a_1)}+\frac{a_1x}{(x+a_1)(x+a_2)}+a_1a_2\frac{x(x+a_4)(x+a_5)...}{(x1+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... }+a_1a_2\frac{a_3(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... } \tag 6$$
$$1=\frac{x}{(x+a_1)}+\frac{a_1x}{(x+a_1)(x+a_2)}+\frac{a_1a_2x}{(x+a_1)(x+a_2)(x+a_3)}+a_1a_2a_3\frac{(x+a_4)(x+a_5)...}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... } \tag 7$$
If we continue in that way to create series then the last term will be :
$$\frac{a_1a_2a_3a_4a_5....}{x(x+a_1)(x+a_2)(x+a_3)(x+a_4)(x+a_5)... }\tag 8$$
The last term must go to zero otherwise the relation I wrote above must be as shown below
$$\frac{1}{x}-\frac{\prod\limits_{j=1}^{\infty}a_j}{x\prod\limits_{j=1}^{\infty} (x+a_j)}=\frac{1}{x+a_1}+\frac{a_1}{(x+a_1)(x+a_2)}+\frac{a_1a_2}{(x+a_1)(x+a_2)(x+a_3)}+\frac{a_1a_2a_3}{(x+a_1)(x+a_2)(x+a_3)(x+a_4)}+... \tag 9$$
$$\frac{1}{x}-\frac{\prod\limits_{j=1}^{\infty}a_j}{x\prod\limits_{j=1}^{\infty} (x+a_j)}=\frac{1}{x+a_1}+\sum\limits_{k=1}^ \infty\frac{\prod\limits_{j=1}^{k}a_j}{\prod\limits_{j=1}^{k+1} (x+a_j)} \tag {10}$$
If $\frac{\prod\limits_{j=1}^{\infty}a_j}{x\prod\limits_{j=1}^{\infty} (x+a_j)}=0$ then the relation will be
$$\frac{1}{x}=\frac{1}{x+a_1}+\sum\limits_{k=1}^ \infty\frac{\prod\limits_{j=1}^{k}a_j}{\prod\limits_{j=1}^{k+1} (x+a_j)} \tag {11}$$
Also The general formula can be written for n terms. $$\frac{1}{x}-\frac{\prod\limits_{j=1}^{n}a_j}{x\prod\limits_{j=1}^{n} (x+a_j)}=\frac{1}{x+a_1}+\sum\limits_{k=1}^ {n-1} \frac{\prod\limits_{j=1}^{k}a_j}{\prod\limits_{j=1}^{k+1} (x+a_j)} \tag {12}$$
I had similiar question about that relation for $x=1$. If you want to see my question and answer, the link
