Question

Chứng tỏ rằng:\(\frac{1}{101}\)+\(\frac{1}{102}\)+\(\frac{1}{103}\)+......…...........+\(\frac{1}{200}\)

>\(\frac{1}{2}\)

Answer 1

Ta có:

\(\frac{1}{101}\)>\(\frac{1}{200}\)

\(\frac{1}{102}\)>\(\frac{1}{200}\)

\(\frac{1}{103}\)>\(\frac{1}{200}\)

...

\(\frac{1}{200}\)=\(\frac{1}{200}\)

\(\frac{1}{101}\)+\(\frac{1}{102}\)+\(\frac{1}{103}\)+...+\(\frac{1}{200}\)>\(\frac{1}{200}\)+\(\frac{1}{200}\)+..+\(\frac{1}{200}\)(100 số hạng)=\(\frac{1}{2}\)

\(\Rightarrow\)\(\frac{1}{101}\)+\(\frac{1}{102}\)+\(\frac{1}{103}\)+...+\(\frac{1}{200}\)>\(\frac{1}{2}\)