Question

Chứng minh rằng với mọi số tự nhiên \(n\ge3:\)

\(B=\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+...+\dfrac{1}{n^3}< \dfrac{1}{12}\)

Answer 1

Ta có: \(\dfrac{1}{3^3}\) < \(\dfrac{1}{2.3.4}\)

\(\dfrac{1}{4^3}\) < \(\dfrac{1}{3.4.5}\)

.......

\(\dfrac{1}{n^3}\) < \(\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

\(\Rightarrow\) \(\dfrac{1}{3^3}\) + \(\dfrac{1}{4^3}\) + ...+ \(\dfrac{1}{n^3}\) < ​\(\dfrac{1}{2.3.4}\)​

+ \(\dfrac{1}{3.4.5}\) + ... + \(\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\) Có:\(\dfrac{1}{2.3.4}\)+ ​\(\dfrac{1}{3.4.5}\)​+...+​\(\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)​ = \(\dfrac{1}{2}\)(\(\dfrac{1}{2.3}\) - \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{3.4}\)- \(\dfrac{1}{4.5}\)+ ... +\(\dfrac{1}{n\left(n-1\right)}\)- \(\dfrac{1}{n}\) + \(\dfrac{1}{n}\) - \(\dfrac{1}{n\left(n+1\right)}\)) = \(\dfrac{1}{2}\)(\(\dfrac{1}{2.3}\) - \(\dfrac{1}{n\left(n+1\right)}\)) = \(\dfrac{1}{12}\)- \(\dfrac{1}{2n\left(n+1\right)}\) < \(\dfrac{1}{12}\) Vậy B = \(\dfrac{1}{3^3}\) + \(\dfrac{1}{4^3}\)+ \(\dfrac{1}{5^3}\)+ ... + \(\dfrac{1}{n^3}\) < \(\dfrac{1}{12}\) Chúc bn học tốt