Question

[Trần Thị Khiêm]

Chung minh 1/2^2+1/3^2+...+1/50^2<1?

Answer 1

[Nguyễn Hải Dương]

Đặt A = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)

Với n \(\in\) N*, n > 1 ta có :

\(\dfrac{1}{n^2}< \dfrac{1}{n\left(n-1\right)}\)( vì 1>0; n² > n(n-1) > 0 )

Áp dụng vào bài ta có :

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

.....

\(\dfrac{1}{50^2}< \dfrac{1}{49.50}\)

=> \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{50^2}\)< \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

=> A < \(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{50-49}{49.50}\)

=> A < \(\dfrac{2}{1.2}-\dfrac{1}{1.2}+\dfrac{3}{2.3}-\dfrac{2}{2.3}+\dfrac{4}{3.4}-\dfrac{3}{3.4}+...+\dfrac{50}{49.50}-\dfrac{49}{49.50}\)

=> A < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

=> A < \(1-\dfrac{1}{50}\) < 1 ( vì \(\dfrac{1}{50}>0\) )

=> A < 1

=> đpcm

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