Chứng minh rằng với mọi số nguyên dương n lớn hơn 3 thì 1/1^3 + 1/2^3 + ... + 1/n^3 < 65/54

Ta có: $n^3 > n^3 - n\quad \forall n \geq 2$

$\to n^3 > (n-1)n(n+1)$

$\to \dfrac{1}{n^3} < \dfrac{1}{(n-1)n(n+1)}$

$\to \dfrac{2}{n^3} < \dfrac{2}{(n-1)n(n+1)}$

$\to \dfrac{2}{n^3} < \dfrac{1}{n(n-1)} - \dfrac{1}{n(n+1)}$

Áp dụng bất đẳng thức vừa chứng minh:

Đặt $S = \dfrac{1}{1^3} + \dfrac{1}{2^3} + \dfrac{1}{3^3} + \dots + \dfrac{1}{n^3}$

$\to 2S = \dfrac{2}{1^3} + \dfrac{2}{2^3} + \dfrac{2}{3^3} + \dots + \dfrac{2}{n^3}$

$\to 2S = 2\cdot\left(\dfrac{1}{1^3} + \dfrac{1}{2^3} +\dfrac{1}{3^3}\right) + \dfrac{1}{3.4} -\dfrac{1}{4.5} + \dots + \dfrac{1}{n(n-1)} - \dfrac{1}{n(n+1)}$

$\to 2S = \dfrac{251}{108} + \dfrac{1}{3.4} - \dfrac{1}{n(n+1)}$

$\to S = \dfrac{65}{54} - \dfrac{1}{2n(n+1)} < \dfrac{65}{54}$

Vậy $\dfrac{1}{1^3} + \dfrac{1}{2^3} + \dfrac{1}{3^3} + \dots + \dfrac{1}{n^3} < \dfrac{65}{54}\quad \forall n \geq 3$