Question

So sánh A = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.........+\dfrac{1}{n^2}\)với 1

Answer 1

k² > k² - 1 = (k-1)(k+1)
⇒ 1/k² < 1/[(k-1).(k+1)] = [1/(k-1) - 1/(k+1)]/2 (*)

Áp dụng (*), ta có:
1/2² + 1/3² + 1/4² + ... + 1/n²
< 1/2² + 1/(2.4) + 1/(3.5) + ... + 1/[(n-1).(n+1)]
= 1/2² + [1/2 - 1/4 + 1/3 - 1/5 + ... + 1/(n-1) - 1/(n+1)]/2
= 1/2² + [1/2 + 1/3 - 1/n - 1/(n+1)]/2
= 2/3 - [1/n + 1/(n+1)]/2 <2/3<1

Answer 2

Vì:\(\dfrac{1}{2^{^2}}>\dfrac{1}{1.2};\dfrac{1}{3^{^2}}>\dfrac{1}{2.3};\dfrac{1}{4^{^2}}>\dfrac{1}{3.4};...;\dfrac{1}{n^{^2}}>\dfrac{1}{\left(n-1\right)n}\)

\(\Rightarrow A>\dfrac{1 }{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)

\(\Rightarrow A>\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(\Rightarrow A>\dfrac{1}{1}-\dfrac{1}{n}\)

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