Sum: $\sum_{n=1}^\infty\prod_{k=1}^n\frac{k}{k+a}=\frac{1}{a-1}$ For the past week, I've been mulling over this Math.SE question. The question was just to prove convergence of 
$$\sum\limits_{n=1}^\infty\frac{n!}{\left(1+\sqrt{2}\right)\left(2+\sqrt{2}\right)\cdots\left(n+\sqrt{2}\right)}$$
but amazingly Mathematica told me it had a remarkably simple closed form: just $1+\sqrt{2}$. After some fiddling, I conjectured for $a>1$:
$$\sum\limits_{n=1}^\infty\prod\limits_{k=1}^n\frac{k}{k+a}=\sum\limits_{n=1}^\infty\frac{n!}{(1+a)(2+a)\cdots(n+a)}=\frac{1}{a-1}$$
I had been quite stuck until today when I saw David H's helpful answer to a similar problem. I have included a solution using the same idea, but I would be interested to know if anyone has another method.

Another possible trick is the following one. For any $a$ such that $\Re(a)>1$, let:
$$ f(a) = \sum_{n=1}^{+\infty}\prod_{k=1}^{n}\frac{k}{k+a} $$
Since:
$$\begin{eqnarray*} \frac{1}{a+1}\, f(a+1) &=& \sum_{n=1}^{+\infty}\frac{n!}{(a+1)(a+2)\cdot\ldots\cdot(a+n+1)}\\&=&\sum_{n=1}^{+\infty}\frac{(n-1)!}{(a+2)\cdot\ldots\cdot(a+n)}\left(\frac{1}{a+1}-\frac{1}{a+n+1}\right)\end{eqnarray*}$$
it follows that:
$$\frac{1}{a+1}\,f(a+1) = f(a+1)-f(a+2)\tag{1} $$
or:
$$ a\, f(a+1) = (a+1)\, f(a+2)\tag{2} $$
So, given that $ z\,f(z+1) $ is an analytic function on $\Re(z)>1$, we have that $z\,f(z+1)$ is constant, and:
$$ 2\, f(3) = \sum_{n=1}^{+\infty}\frac{12}{(n+1)(n+2)(n+3)}=\sum_{n=1}^{+\infty}\left(\frac{6}{n+1}-\frac{12}{n+2}+\frac{6}{n+3}\right)=\color{red}{1}\tag{3} $$
by the telescopic property. This gives that over $\Re(a)>1$,
$$ f(a) = \color{red}{\frac{1}{a-1}}\tag{4}$$
holds.
