Xét vế phải\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{199}-\frac{1}{200}\)

=\(\left(1+\frac{1}{3}+\frac{1}{5}+..+\frac{1}{199}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{200}\right)\)

=\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}\right)-2.\left(\frac{1}{2}-\frac{1}{4}-...-\frac{1}{200}\right)\)

=\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-...-\frac{1}{100}\)

=\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)

Xét vế phải 1 − 1 2 + 1 3 − 1 4 + . . . . . + 1 199 − 1 200 1− 2 1 ​ + 3 1 ​ − 4 1 ​ +.....+ 199 1 ​ − 200 1 ​ = ( 1 + 1 3 + 1 5 + . . + 1 199 ) − ( 1 2 + 1 4 + . . . + 1 200 ) (1+ 3 1 ​ + 5 1 ​ +..+ 199 1 ​ )−( 2 1 ​ + 4 1 ​ +...+ 200 1 ​ ) = ( 1 + 1 2 + 1 3 + 1 4 + 1 5 + . . . + 1 199 + 1 200 ) − 2. ( 1 2 − 1 4 − . . . − 1 200 ) (1+ 2 1 ​ + 3 1 ​ + 4 1 ​ + 5 1 ​ +...+ 199 1 ​ + 200 1 ​ )−2.( 2 1 ​ − 4 1 ​ −...− 200 1 ​ ) = 1 + 1 2 + 1 3 + 1 4 + 1 5 + . . . + 1 199 + 1 200 − 1 − 1 2 − . . . − 1 100 1+ 2 1 ​ + 3 1 ​ + 4 1 ​ + 5 1 ​ +...+ 199 1 ​ + 200 1 ​ −1− 2 1 ​ −...− 100 1 ​ = 1 101 + 1 102 + . . . + 1 200 101 1 ​ + 102 1 ​ +...+ 200 1 ​

Ta có:  \(\dfrac{1}{101}>\dfrac{1}{200}\)

Tương tự ta có: \(\dfrac{1}{102}>\dfrac{1}{200}\) ;....; \(\dfrac{1}{199}>\dfrac{1}{200}\)

\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{1}{200}.100\)

\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{100}{200}\)

\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{1}{2}\left(đpcm\right)\)