Question

[Nguyễn Bùi Đại Hiệp]

Chứng minh rằng với mọi n\(\ge2\)ta có

\(\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{n^3}< \frac{1}{4}\)

Answer 1

[Akai Haruma]

Lời giải:

Xét số hạng tổng quát \(\frac{1}{n^3}\)

\((n-1)(n+1)=n^2-1< n^2\)

\(\Rightarrow (n-1)n(n+1)< n^3\)

\(\Rightarrow \frac{1}{(n-1)n(n+1)}>\frac{1}{n^3}\)

Thay $n=2,3,4,.....$. Khi đó ta có:

\(\frac{1}{2^3}+\frac{1}{3^3}+....+\frac{1}{n^3}<\underbrace{ \frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{(n-1)n(n+1)}}_{A}(*)\)

Mà:

\(2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+....+\frac{(n+1)-(n-1)}{(n-1)n(n+1)}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{(n-1)n}-\frac{1}{n(n+1)}\)

\(=\frac{1}{2}-\frac{1}{n(n+1)}< \frac{1}{2}\)

\(\Rightarrow A< \frac{1}{4}(**)\)

Từ \((*) ;(**)\Rightarrow \frac{1}{2^3}+\frac{1}{3^3}+....+\frac{1}{n^3}< \frac{1}{4}\)

Ta có đpcm.