Question

Cho A= 1/2² + 1/3² +.......+ 1/2005²

        B= 1/1.2 + 1/2.3 +....+ 1/2004.2005

a/So sánh A với B

b/Chứng minh A < 1

Answer 1

a) A= 1/2² +1/3²+...+1/2005²

A= 1/2.2 + 1/3.3 +....+1/2005.2005

Vì 1/2.2 < 1/1.2 ; 1/3.3 < 1/6;.....; 1/2005.2005 < 1/2004.2005 nên A= 1/2² +1/3²+...+1/2005² < 1/1.2 + 1/2.3 +....+ 1/2004.2005

=> A < B

Vậy...

Answer 2

a) \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2005^2}\)

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

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

...

\(\frac{1}{2005^2}=\frac{1}{2005\cdot2005}< \frac{1}{2004\cdot2005}\)

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2005^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2004\cdot2005}=B\)

\(\Rightarrow A< B\)

b) \(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2004\cdot2005}=\frac{1}{1}-\frac{1}{2005}=\frac{2004}{2005}< 1\)

Theo câu a) => \(A< B< 1\)

=> A < 1 ( đpcm )