Question

chứng minh rằng:

\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}\)

Answer 1

Ta thấy: k² > (k - 1)(k + 1)

Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(< \dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)

\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right).\dfrac{1}{2}\)

\(=\left(1-\dfrac{1}{101}\right).\dfrac{1}{2}\)

\(=\dfrac{100}{101}.\dfrac{1}{2}< 1.\dfrac{1}{2}=\dfrac{1}{2}\)