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.

The idea of this solution is to appeal to the Beta function and then to exchange the order of integration and summation (made possible by Fubini's theorem).
$$\begin{align}
\sum\limits_{n=1}^\infty\prod\limits_{k=1}^n\frac{k}{k+a}&=\sum\limits_{n=1}^\infty\frac{n!}{(1+a)(2+a)\cdots(n+a)}
\\&=\sum\limits_{n=1}^\infty\frac{\Gamma(n+1)\Gamma(1+a)}{\Gamma(n+a+1)}
\\&=\Gamma(1+a)\sum\limits_{n=1}^\infty\frac{\Gamma(n+1)}{\Gamma(n+a+1)}
\\&=\frac{\Gamma(1+a)}{\Gamma(a)}\sum\limits_{n=1}^\infty\frac{\Gamma(n+1)\Gamma(a)}{\Gamma(n+a+1)}
\\&=a\sum\limits_{n=1}^\infty \operatorname{B}(n+1,a)
\\&=a\sum\limits_{n=1}^\infty\int\limits_0^1t^n(1-t)^{a-1}\,dt
\\&=a\int\limits_0^1\sum\limits_{n=1}^\infty t^n(1-t)^{a-1}\,dt
\\&=a\int\limits_0^1\frac{t(1-t)^{a-1}}{1-t}\,dt
\\&=a\int\limits_0^1t(1-t)^{a-2}\,dt
\\&=a\operatorname{B}(2,a-1)
\\&=\frac{a}{a(a-1)}
\\&=\frac{1}{a-1}
\end{align}$$
Note that we used $a>1$ when evaluating $\operatorname{B}(2,a-1)$, since the beta function is only defined when both arguments are greater than $0$.
A final note: the restriction $a>1$ is sharp in the sense that when $a=1$ the inside product simplifies to $$\prod\limits_{k=1}^n\frac{k}{k+1}=\frac{1}{n+1}$$
so the sum becomes
$$\sum\limits_{n=1}^\infty\prod\limits_{k=1}^n\frac{k}{k+1}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots=\infty$$
and for $a<1$ in essence we can use the comparison test with the $a=1$ case to show that the sum diverges.
