UNSW MathSoc Championship Q11

We are told that we have to compute $\prod_{k=4}^{\infty} \frac{2k^2 -5k - 3}{2k^2 - 5k + 2}$ . First, let's factorize these polynomials: $2k^2 -5k - 3 = (k-3)(2k + 1)$ and $2k^2 - 5k + 2 = (2k-1)(k-2)$ to write the product as $\prod_{k=4}^{\infty} \frac{(2k+1)(k-3)}{(2k-1)(k-2)}$ . Next, the substitution $n=k-3$ makes this equal to $\prod_{n=1}^{\infty} \frac{(2n + 7)n}{(2n + 5)(n+1)}$ . In the previous step one would hope that something would cancel but it didn't so we are forced to exploit the factorization in some other way:

Let $P_M = \prod_{n=1}^{M} \frac{(2n + 7)n}{(2n + 5)(n+1)}$ . These are the partial products. We may rewrite this as $\prod_{n=1}^M \frac{n}{n+1} \prod_{n=1}^M \frac{2n+7}{2n+5}$ . First notice that $\prod_{n=1}^M \frac{n}{n+1} = \frac{1}{M+1}$ and $\prod_{n=1}^M \frac{2n+7}{2n+5} = \frac{2M + 7}{7}$ . This is because as you keep multiplying terms, past terms cancel. So $P_M = \frac{2M + 7}{7M + 7}$ and we may compute $\lim_{M \to \infty} P_M = \frac{2}{7}$ .