Question

So sánh :

B = 1/2²+ 1/3²+ 1/4²+ ... +1/100² với 1

Answer 1

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\) với 1

\(B=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{1}-\dfrac{1}{100}\)

\(=\dfrac{99}{100}< 1\)

=> \(B< 1\)

Answer 2

Ta thấy

\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)

B < \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

B < \(\left(1-\dfrac{1}{100}\right)\)

B < \(\dfrac{99}{100}\)

Vì \(\dfrac{99}{100}< 1\)

\(\Rightarrow B< 1\)