Ta có : D = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}=\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{10.10}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}< 1\)
=> D < 1 (đpcm)
Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^3}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
...
\(\frac{1}{10^2}< \frac{1}{9.10}\)
=)) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
Mà \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}=\frac{9}{10}< 1\)
=)) A < 1 (đpcm)
D à, lỗi @@ làm tưởng A chứ, sửa KL nha eiu
\(D=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\); \(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\); ... ; \(\frac{1}{10^2}=\frac{1}{10\cdot10}< \frac{1}{9\cdot10}\)
Cộng theo vế ta có :
\(D=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\)
\(D< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(D< \frac{1}{1}-\frac{1}{10}=\frac{9}{10}\)(1)
Lại có \(\frac{9}{10}< 1\)(2)
Từ (1) và (2) => \(D< \frac{9}{10}< 1\Rightarrow D< 1\left(đpcm\right)\)