KD

Cho S=1/5^2+1/7^2+1/9^2+...+1/103^2 chứng minh S < 5/32

YN
10 tháng 2 2023 lúc 19:57

\(S=\dfrac{1}{5^2}+\dfrac{1}{7^2}+\dfrac{1}{9^2}+...+\dfrac{1}{103^2}\)

\(\Rightarrow2S=\dfrac{2}{5^2}+\dfrac{2}{7^2}+\dfrac{2}{9^2}+...+\dfrac{2}{103^2}\)

Có:

\(\dfrac{2}{5^2}=\dfrac{2}{5.5}< \dfrac{2}{4.6}=\dfrac{1}{4}-\dfrac{1}{6}\)

\(\dfrac{2}{7^2}=\dfrac{2}{7.7}< \dfrac{2}{6.8}=\dfrac{1}{6}-\dfrac{1}{8}\)

\(\dfrac{2}{9^2}=\dfrac{2}{9.9}< \dfrac{2}{8.10}=\dfrac{1}{8}-\dfrac{1}{10}\)

...

\(\dfrac{2}{103^2}=\dfrac{2}{103.103}< \dfrac{1}{102.104}=\dfrac{1}{102}-\dfrac{1}{104}\)

\(\Rightarrow2S< \dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+...+\dfrac{1}{102}-\dfrac{1}{104}\)

\(\Rightarrow2S< \dfrac{25}{104}\)

\(\Rightarrow S< \dfrac{25}{208}< \dfrac{5}{32}\)

\(\Rightarrow S< \dfrac{5}{32}\).

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H24
10 tháng 2 2023 lúc 19:57

Ta có:
\(\dfrac{1}{5^2}< \dfrac{1}{4.6}\)
\(\dfrac{1}{7^2}< \dfrac{1}{6.8}\)
\(\dfrac{1}{9^2}< \dfrac{1}{8.10}\)
\(...\)
\(\dfrac{1}{103^2}< \dfrac{1}{102.104}\)
\(\Rightarrow S\)\(< \dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}+...+\dfrac{1}{102.104}\)\(\left(1\right)\)
Đặt \(A=\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}+...+\dfrac{1}{102.104}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{4.6}+\dfrac{2}{6.8}+\dfrac{2}{8.10}+...+\dfrac{2}{102.104}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+...+\dfrac{1}{102}-\dfrac{1}{104}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{4}-\dfrac{1}{104}\right)\)
\(=\dfrac{1}{2}.\dfrac{25}{104}\)
\(=\dfrac{25}{208}< \dfrac{25}{160}\)\(\left(2\right)\)
Mà \(\dfrac{25}{160}=\dfrac{5}{32}\)\(\left(3\right)\)
Từ \(\left(1\right),\left(2\right)\) và \(\left(3\right)\)
\(\Rightarrow S< \dfrac{5}{32}\)

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