limit pochhammer symbol $\lim_{n \rightarrow \infty} \left(\frac{1}{\sqrt{n}},\frac{1}{\sqrt{n}}\right)_n= 1$

Question

I am looking for a proof of the following relation:

$$\lim_{n \rightarrow \infty} \left(\frac{1}{\sqrt{n}},\frac{1}{\sqrt{n}} \right)_n= \lim_{n \rightarrow \infty} \prod_{i=1}^n \left( 1 - \left(\frac{1}{\sqrt{n}} \right)^i \right)= 1 $$

This involves the limit of a Pochhammer symbol. Any reference is highly appreciated

Answer 1

The following result can be applied to the OP's problem:

Theorem Let $\{c_{n,m}:1\leq m\leq m_n\}\subset\mathbb{C}$. Suppose that

1. $\lim\limits_{n\rightarrow0}\sup_{1\leq m\leq m_n}|c_{n,m}|=0$,
2. $\lim\limits_{n\rightarrow\infty}\sum^{m_n}_{m=1}c_{n,m}=c\in\mathbb{C}$,
3. and $M:=\sup_n\sum^{m_n}_{m=1}|c_{n,m}|<\infty$. Then \begin{align}\prod^{m_n}_{m=1}(1+c_{n,m})=e^c \end{align}

In the OP's case, set $c_{n, m}=-n^{-m/2}$, $1\leq m\leq n$, $n\in\mathbb{N}$.

This theorem appears in many graduate textbooks in probability (Durrett's or Breiman's for example). A proof is also discussed in MSE here along with more examples of applications.