I wonder if there is an elementary way to show
$$\sum_{n=0}^{\infty} \frac{a(a+1) \ldots (a+n)}{b(b+1) \ldots (b+n)}=\frac{a}{b-a-1}$$
if $b > a +1, a > 0$ .I'm reading Stefan Hildebrandt Analysis 1 and one exercise is to prove that. I've seen a proof using the Gamma function but until that point in the book only basic convergence theorems for sequences and series were presented so I wonder if there is some smart way to go about it using more elementary methods as I couldn't find it.
Thanks.
This is a simple consequence of the following identity:
$$ \begin{align*} \sum_{n=0}^{N} \frac{a(a+1)\cdots(a+n)}{b(b+1)\cdots(b+n)} &= \frac{1}{b-a-1} \left[ a - \frac{a(a+1) \cdots (a+N+1)}{b(b+1) \cdots (b+N)} \right] \end{align*} $$
This is easily proved by induction on $N$, so I will skip the proof. (Remark. This can be thought of as a generalized version of the hockey-stick identity in disguise.) Now if $b > a+1 > 0$, then by invoking the inequality $1-x \leq e^{-x}$,
$$ \begin{align*} \frac{a(a+1) \cdots (a+N+1)}{b(b+1) \cdots (b+N)} &= a \prod_{k=0}^{N} \left( 1 - \frac{b-a-1}{b+k} \right) \\ &\leq a \exp \left( - \sum_{k=0}^{N} \frac{b-a-1}{b+k} \right). \end{align*} $$
Since the series $\sum_{k=0}^{\infty} \frac{b-a-1}{b+k}$ diverges to $+\infty$, it follows that
$$ \frac{a(a+1) \cdots (a+N+1)}{b(b+1) \cdots (b+N)} \to 0 $$
as $N \to \infty$ and the desired identity follows.
