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.
 A: 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.
A: Here is a completely elementary proof, which only needs introductory calculus concepts:
$$a_{n+1} = \frac{(n+1)!}{(a+1)(a+2)\dots(a+n+1)} =  \frac{1}{a-1}\left(\frac{(n+1)!}{(a+1)(a+2) \dots (a+n)}  -\frac{(n+2)!}{(a+1)(a+2) \dots (a+n+1)}\right) = b_{n+1} - b_{n+2}$$
where $a_n$ is the term of our series and $$b_n = \frac{1}{a-1}\left(\frac{n!}{(a+1)(a+2) \dots (a+n-1)}\right)$$ with $b_1 = \frac{1}{a-1}$
Thus the sum we seek, telescopes!
Giving us
$$\sum_{k=1}^{n} a_k = \sum_{k=1}^{n} (b_{k} - b_{k+1}) = b_1 - b_{n+1}$$
Thus we just need to compute $\lim b_n$ 
(we basically need to show that $b_n \to 0$ to match the limit being $\frac{1}{a-1}$)
Now we have that $a \gt 1$, so let $a = 1 + x$ for $x \gt 0$.
$$ (a-1)b_n =\frac{n!}{(a+1)(a+2) \dots (a+n-1)} = \frac{1}{(1 + x/2)(1+x/3)\dots(1+x/n)} $$
Let $ M = \lceil x \rceil$ and consider the product
$$p_n  = \prod_{k=M}^{n} \left(1 + \frac{x}{k}\right)$$
It is enough to show that $\log p_n \to \infty$ (that proves that $b_n \to 0$).
Now we have that $\dfrac{1}{1+t} \gt 1-t$ for $0 \lt t \le 1$ and thus integrating between $0$ and $y$ (where $y \le 1$) we get that
$$ \log(1+y) \ge y - \dfrac{y^2}{2}$$
Now $$\log p_n = \sum_{k=M}^{n} \log (1 + \frac{x}{k})$$
$$ \ge \sum_{k=M}^{n} (\frac{x}{k} - \frac{x^2}{2k^2})$$
This goes to $\infty$ as the harmonic series diverges, and the sum of reciprocals of the squares converges.
Thus the sum of your sequence is $$b_1 = \frac{1}{a-1}$$
A: 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.
