Ta chứng minh với \(\hept{\begin{cases}n\ge a+2\\a\ge1\end{cases}}\)thì
\(\frac{1}{a}+\frac{1}{n}>\frac{1}{a+1}+\frac{1}{n-1}\)
\(\Leftrightarrow\frac{a+n}{an}>\frac{a+n}{an-a+n-1}\)
\(\Leftrightarrow an< an-a+n-1\)
\(\Leftrightarrow n>a+1\)(đúng)
Từ đó ta có
\(\frac{1}{2018}+\frac{1}{6052}>\frac{1}{2019}+\frac{1}{6051}>...>\frac{1}{4034}+\frac{1}{4036}>\frac{1}{4035}+\frac{1}{4035}=\frac{2}{4035}\) (có 2017 nhóm lớn hơn \(\frac{2}{4035}\) tất cả)
\(\Rightarrow S=\frac{1}{2017+1}+\frac{1}{2017+2}+...+\frac{1}{3.2017+1}=\frac{1}{2018}+\frac{1}{2019}+...+\frac{1}{6052}\)
\(>\frac{2}{4035}+\frac{2}{4035}+...+\frac{2}{4035}+\frac{1}{4035}=\frac{2017.2}{4035}+\frac{1}{4035}=\frac{4035}{4035}=1\)
