Question

chứng minh S=\(\frac{1}{2017+1}+\frac{1}{2017+2}+...+\frac{1}{3\cdot2017+1}>1.\)

mk cần gấp nhanh nha 3 tick đó

Answer 1

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)

\(\Rightarrow S=\frac{1}{2017+1}+\frac{1}{2017+2}+...+\frac{1}{3.2017+1}=\frac{1}{2018}+\frac{1}{2019}+...+\frac{1}{6052}\)

\(>\frac{2}{4035}+\frac{2}{4035}+...+\frac{2}{4035}+\frac{1}{4035}=\frac{2017.2}{4035}+\frac{1}{4035}=\frac{4035}{4035}=1\)