16 $\frac{1}{e_{1}}=\frac{1}{2},\frac{1}{e_{1}}+\frac{1}{e_{2}}=\frac{5}{6},\frac{1}{e_{1}}+\frac{1}{e_{2}}+\frac{1}{e_{3}}=\frac{41}{42}$，由此猜测$\sum_{i=1}^{k}\frac{1}{e_{i}}=\frac{e_{k+1}-2}{e_{k+1}-1}$
假设前$n$项都成立，即$\sum_{i=1}^{n}\frac{1}{e_{i}}=\frac{e_{n+1}-2}{e_{n+1}-1}$
那么$\sum_{i=1}^{n+1}\frac{1}{e_{i}}=\frac{e_{n+1}-2}{e_{n+1}-1}+\frac{1}{e_{n+1}}=\frac{(e_{n+1}-1)e_{n+1}-1}{(e_{n+1}-1)e_{n+1}}=\frac{e_{n+2}-2}{e_{n+2}-1}$
17 $Gcd(f_{m},f_{n})=Gcd(f_{m},(f_{n})mod(f_{m}))=Gcd(f_{m},2)=1$