jmc

We have that $$\frac{\left(\genfrac{}{}{0ex}{}{n}{15}\right)}{\left(\genfrac{}{}{0ex}{}{n}{17}\right)}=\frac{\frac{n!}{15!(n-15)!}}{\frac{n!}{17!(n-17!)}}=\frac{17!(n-17)!}{15!(n-15)!}=\frac{17\cdot 16}{(n-15)(n-16)}.$$By partial fractions, $$\frac{1}{(n-15)(n-16)}=\frac{1}{n-16}-\frac{1}{n-15}.$$We can also observe that $$\frac{1}{(n-15)(n-16)}=\frac{(n-15)-(n-16)}{(n-15)(n-16)}=\frac{1}{n-16}-\frac{1}{n-15}.$$Therefore, $$\begin{array}{rl}\sum _{n=17}^{\infty }\frac{\left(\genfrac{}{}{0ex}{}{n}{15}\right)}{\left(\genfrac{}{}{0ex}{}{n}{17}\right)}& =272\sum _{n=17}^{\infty }\frac{1}{(n-15)(n-16)}\\ & =272\sum _{n=17}^{\infty }\left(\frac{1}{n-16}-\frac{1}{n-15}\right)\\ & =272\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots \right]\\ & =\boxed{272}.\end{array}$$