Question

CMR n\(\in \)N, n>3

\(\frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{n^3} <2 \)

@Akai Haruma

Answer 1

Với \(n\ge3\) thì ta có:

\(\dfrac{1}{n^3}< \dfrac{1}{\left(n-2\right)\left(n-1\right)n}=\dfrac{1}{2}\left(\dfrac{1}{\left(n-2\right)\left(n-1\right)}-\dfrac{1}{\left(n-1\right)n}\right)\)

Áp dụng vào bài toán ta được

\(\dfrac{1}{1^3}+\dfrac{1}{2^3}+...+\dfrac{1}{n^3}\)

\(< 1+\dfrac{1}{8}+\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-2\right)\left(n-1\right)}-\dfrac{1}{\left(n-1\right)n}\right)\)

\(=1+\dfrac{1}{8}+\dfrac{1}{2}.\left(\dfrac{1}{2}-\dfrac{1}{\left(n-1\right)n}\right)\)

\(< 1+\dfrac{1}{8}+\dfrac{1}{4}=\dfrac{11}{8}< 2\)