\(\) Chứng minh: a) \(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}< 2\) b) \(\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}-\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}-\dfrac{1}{3^{100}}< \dfrac{1}{4}\)
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a: Đặt \(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}=\dfrac{1}{99}-\dfrac{1}{100}\)

Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< 1-\dfrac{1}{100}< 1\)

=>\(A=1+\dfrac{1}{2^2}+...+\dfrac{1}{100^2}< 1+1=2\)

b: Đặt \(B=\dfrac{1}{3}-\dfrac{1}{3^2}+\dfrac{1}{3^3}-...-\dfrac{1}{3^{100}}\)

=>\(3B=1-\dfrac{1}{3}+\dfrac{1}{3^2}-...-\dfrac{1}{3^{99}}\)

=>\(3B+B=1-\dfrac{1}{3}+\dfrac{1}{3^2}-...-\dfrac{1}{3^{99}}+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{100}}\)

=>\(4B=1-\dfrac{1}{3^{100}}\)

=>\(B=\dfrac{1}{4}-\dfrac{1}{4\cdot3^{100}}< \dfrac{1}{4}\)