Convergence of an infinite product

Remarks

The value does not depend on the choice of $m$. If the product converges, then $a_n \to 1$, since $a_n = q_n / q_{n-1}$ for $n > m$. Example: $\prod_{n=2}^{\infty} (1 - 1/n^2)$ converges with value $\tfrac12$. With $m = 2$, the partial products are $\prod_{n=2}^{N} \frac{(n-1)(n+1)}{n^2} = \frac{N+1}{2N}$, which tend to $\tfrac12 \ne 0$. Non-example: $\prod_{n=2}^{\infty} (1 - 1/n)$ has partial products $1/N \to 0$, so it fails clause 3 for every $m$ and diverges. The product $0 \cdot 2 \cdot 2 \cdot 2 \cdots$ also diverges, because its tail products are growing powers of $2$ with no limit.