sequences and series - Sum: $sum_{n=1}^inftyprod_{k=1}^nfrac{k}{k+a}=frac{1}{a-1}$

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.

Answer

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 $$b_1 = \frac{1}{a-1}$$