calculus - Limit of a convergent series

For a research project, after some manipulation I come up with a convergent series that I have to prove its limit. The statement is the following:

$ \lim_{n \rightarrow \infty } \displaystyle \sum_{k=1}^n \left[ \prod_{j=n-k+1}^\infty (1-q^j)-1 \right]q^k = 0, \; 0

The LHS can be rewritten by the q-shifted factorial: $(q,q)_n = \prod_{i=1}^n(1-q^i)$ as follows:

$\lim_{n \rightarrow \infty} \displaystyle \sum_{k=1}^n \left[ \frac{(q,q)_\infty}{(q,q)_{n-k}} -1 \right] q^k = 0, \; 0

In fact the second statement is the original limit that I have to prove. I tested the limit numerically by varying $n$, the series seem to converge pretty fast. However, I couldn't figure out how to prove it analytically. If anyone can have an idea to share, I would be grateful.

Answer

In your sums, call the terms in brackets $c(k,n).$ Let's think of $c(k,n)$ as defined for all $k,n\in \mathbb {N},$ with $c(k,n) = 0$ for $k>n.$ So we are dealing with $\sum_{k=1}^{\infty}c(k,n)q^k.$ Now $|c(k,n)|\le 2$ for all $k,n.$ And for fixed $k,\lim_{n\to \infty}c(k,n) = 0.$ Because $\sum_{k=1}^{\infty}q^k < \infty,$ the dominated convergence theorem tells you

$$\lim_{n\to\infty}\sum_{k=1}^{\infty}c(k,n)q^k = \sum_{k=1}^{\infty}[\lim_{n\to\infty}c(k,n)q^k]=0.$$