Chứng minh rằng :a) S=\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+...+\(\frac{3}{n(n+3)}\)< 1 .b) Cho : S=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{9^2}\). Chứng minh :\(\frac{2}{5}\)< S <\(\frac{8}{9}\). Gợi ý : Chứng minh \(\frac{1}{b}\)-\(\frac{1}{b+1}\)<\(\frac{1}{b^2}\)<\(\frac{1}{b-1}\)-\(\frac{1}{b}\) ( b\(\varepsilon\)\(ℕ\), b > 1 )...
Đọc tiếp
Chứng minh rằng :
a) S=\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+...+\(\frac{3}{n(n+3)}\)< 1 .
b) Cho : S=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{9^2}\). Chứng minh :\(\frac{2}{5}\)< S <\(\frac{8}{9}\).
Gợi ý : Chứng minh \(\frac{1}{b}\)-\(\frac{1}{b+1}\)<\(\frac{1}{b^2}\)<\(\frac{1}{b-1}\)-\(\frac{1}{b}\) ( b\(\varepsilon\)\(ℕ\), b > 1 ) .
a)Ta có: \(\frac{3}{1.4}=\frac{4-1}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{7-4}{4.7}=\frac{1}{4}-\frac{1}{7}\)
... . . . .
\(\frac{3}{n\left(n+3\right)}=\frac{1}{n}-\frac{1}{n+3}\)
\(\Leftrightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+3}< 1^{\left(đpcm\right)}\)
b) Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Suy ra \(\frac{2}{5}< S\) (1)
Ta lại có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\)
Từ đó suy ra S < 8/9
Từ (1) và (2) suy ra đpcm