TITLE: How to prove that $ \sum\limits_{n=1}^{\infty}n\prod\limits_{k=1}^{n}\frac{1}{1+ka} = \frac{1}{a} $?
QUESTION [15 upvotes]: Mathematica tells me that
$$
\sum_{n=1}^{\infty}n\prod_{k=1}^{n}\frac{1}{1+ka} = \frac{1}{a}
$$
I could prove it for $a\rightarrow 0$, $a=1$ and $a\rightarrow \infty$, but could not find a general proof for real $a>0$. Can anyone find one? Many thanks!

REPLY [9 votes]: This may be seen as a simple heads-and-tails game in disguise...

Consider a heads-and-tails game such that the probability of success of the $n$th throw is $$p_n=\frac{na}{1+na}.$$ Then the probability that the $n$th throw is the first success is
$$
p_n\prod_{k=1}^{n-1}(1-p_k)=na\prod_{k=1}^n\frac1{1+ka},
$$
hence $$
\sum_{n=1}^{\infty}n\prod_{k=1}^{n}\frac{1}{1+ka}=\frac1a\,P(A),
$$
where $A$ is the event that at least one success happens, ever. The event that there is no success amongst the $n$ first throws has probability
$$
\prod_{k=1}^n(1-p_k)=\prod_{k=1}^n\frac1{1+ka},
$$
which converges to $0$ hence $P(A)=1$, QED.
Nota: This interpretation also yields the finitary identity
$$\sum_{n=1}^Nn\prod_{k=1}^{n}\frac{1}{1+ka}=\frac1a\,\left(1-\prod_{n=1}^{N}\frac{1}{1+na}\right),
$$
and a natural interpretation of it, since, on the one hand, $a$ times the LHS is the probability that the first success is seen amongst the $N$ first throws, and, on the other hand, the probability of the complementary event is, by independence, the product in the RHS.