Question

Chứng minh:

S\(_2\)= \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^1}\)+...+\(\dfrac{1}{50^2}\)<1

Answer 1

Sửa đề: \(S_2=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{50^2}\)

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{50^2}<\frac{1}{49\cdot50}=\frac{1}{49}-\frac{1}{50}\)

Do đó: \(S_2=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{50^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{49}-\frac{1}{50}\)

=>\(S_2\) <1(ĐPCM)