How to prove that $ \sum\limits_{n=1}^{\infty}n\prod\limits_{k=1}^{n}\frac{1}{1+ka} = \frac{1}{a} $? 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! 
 A: Lemma:
\begin{align}
\sum_{n=1}^{N} \frac{n}{\prod_{k=1}^n(1+ka)}=\frac{1}{a}\left[1-\frac{1}{\prod_{k=1}^N(1+ka)}\right].
\end{align}
Proof: Induction on $N$.
A: 
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.
A: We have that:
$$A_n=\prod_{k=0}^n\frac{1}{1+ka} =\frac{1}{n!}\prod_{k=1}^{n}\frac{1}{a+\frac{1}{k}},$$
that can be regarded as a meromorphic function of the variable $a$. Moreover, for any $j\in[1,n]$:
$$\begin{eqnarray*}\operatorname{Res}\left(A_n,a=-\frac{1}{j}\right)&=&\frac{1}{n!}\prod_{\substack{k=1\\k\neq j}}^{n}\frac{1}{-\frac{1}{j}+\frac{1}{k}}=(-1)^n j^{n-2}\prod_{\substack{k=1\\k\neq j}}^{n}\frac{1}{k-j}\\&=&\frac{(-1)^n\,j^{n-2}}{(-1)^{j-1}(j-1)!(n-j)!}\\&=&-\frac{(-1)^{n-j}\, j^{n-1}}{n!}\binom{n}{j}\end{eqnarray*} $$
hence the residue of $\sum_{n=1}^{+\infty}n A_n$ in $a=-\frac{1}{j}$ is:
$$\begin{eqnarray*}-\sum_{n\geq j}\frac{(-1)^{n-j}\, j^{n-1}}{(n-1)!}\binom{n}{j}&=&-j!\cdot\frac{d^j}{dx^j}\left.\sum_{n\geq j}\frac{(-1)^{n-j}\, j^{n-1} x^n}{(n-1)!}\right|_{x=1}\\&=&-j!\cdot\frac{d^j}{dx^j}\left.\sum_{n=0}^{+\infty}\frac{(-1)^{n-j}\, j^{n-1} x^n}{(n-1)!}\right|_{x=1}\\&=&j!\cdot(-1)^j\frac{d^j}{dx^j}\left.\left(x\,e^{-jx}\right)\right|_{x=1}=0,\end{eqnarray*}$$
and the calculations you made for the cases $a=1$ and $a\to 0$ are enough to state 
$$\sum_{n=1}^{+\infty} n A_n = \frac{1}{a}$$
for any $a>0$. In another fashion, using the Euler Beta function:
$$\begin{eqnarray*}\sum_{n=1}^{+\infty}n A_n &=& \sum_{n\geq 1}\frac{n}{a^n}\cdot\frac{\Gamma(1+1/a)}{\Gamma(1+1/a+n)}=\int_{0}^{1}(1-x)^{1/a}\sum_{n\geq 1}\frac{n x^{n-1}}{a^n \Gamma(n)}dx\\&=&\int_{0}^{1}(1-x)^{1/a}e^{x/a}\frac{x+a}{a^2}dx\\&=&\int_{0}^{1}(1-x)^{1/a}\frac{d}{dx}\left(\frac{x}{a}e^{x/a}\right)dx\\&=&\frac{1}{a}\int_{0}^{1}(1-x)^{1/a-1}\frac{x}{a}e^{x/a}dx\\&=&\frac{1}{a}\int_{0}^{1}(1-x^a)\exp\left(\frac{1-x^a}{a}\right)dx=\frac{1}{a},\end{eqnarray*}$$
since we have a cancellation induced by integration by parts:
$$\begin{eqnarray*}\int_{0}^{1}x^a \exp\left(\frac{1-x^a}{a}\right)dx&=&\int_{0}^{1}x\cdot\left(x^{a-1}\exp\left(\frac{1-x^a}{a}\right)\right)dx\\&=&\left.-x\exp\left(\frac{1-x^a}{a}\right)\right|_{0}^{1}+\int_{0}^{1} \exp\left(\frac{1-x^a}{a}\right)dx.\end{eqnarray*}$$
