jmc

For the sum $b_n,$ let $j=n-k,$ so $k=n-j.$ Then $$\begin{array}{rl}{b}_n& =\sum _{k=0}^n\frac{k}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}\\ & =\sum _{j=n}^0\frac{n-j}{\left(\genfrac{}{}{0ex}{}{n}{n-j}\right)}\\ & =\sum _{j=0}^n\frac{n-j}{\left(\genfrac{}{}{0ex}{}{n}{j}\right)}\\ & =\sum _{k=0}^n\frac{n-k}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)},\end{array}$$so $$b_n+b_n=\sum _{k=0}^n\frac{k}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}+\sum _{k=0}^n\frac{n-k}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}=\sum _{k=0}^n\frac{n}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}=n\sum _{k=0}^n\frac{1}{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}=n{a}_n.$$Then $2b_n=n{a}_n,$ so $\frac{a_n}{b_n}=\boxed{\frac{2}{n}}.$