Đề bài

Chứng minh rằng

 A = \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{10}^2}}} < 1\)

Phương pháp giải

- Sử dụng kiến thức nếu A < B và B < C thì ta suy ra A < C

- Ta có 

\(\frac{1}{{{2^2}}} = \frac{1}{{2.2}} < \frac{1}{{1.2}}\)

\(\frac{1}{{{3^2}}} = \frac{1}{{3.3}} < \frac{1}{{2.3}}\)

Nên suy ra \(A < \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{9.10}}\)

Tính tổng \(\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{9.10}}\) bằng cách tách:

\(\begin{array}{l}\frac{1}{{1.2}} = 1 - \frac{1}{2}\\\frac{1}{{2.3}} = \frac{1}{2} - \frac{1}{3}\end{array}\)

...

Lời giải của GV Loigiaihay.com

Chứng minh rằng

 A = \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{10}^2}}} < 1\)

Ta có: \(\frac{1}{{{2^2}}} = \frac{1}{{2.2}} < \frac{1}{{1.2}}\)

\(\frac{1}{{{3^2}}} = \frac{1}{{3.3}} < \frac{1}{{2.3}}\)

\(\frac{1}{{{4^2}}} = \frac{1}{{4.4}} < \frac{1}{{3.4}}\)

…

\(\frac{1}{{{{10}^2}}} = \frac{1}{{10.10}} < \frac{1}{{9.10}}\)

Nên \(\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{10}^2}}} < \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{9.10}}\)

A <\(\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{9.10}}\)

Ta lại có: \(\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{9.10}}\) = 1 - \(\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{9} - \frac{1}{{10}} = 1 - \frac{1}{{10}} = \frac{9}{{10}}\)

Vì \(\frac{9}{{10}} < 1\) nên A < 1