Xét tính bị chặn của dãy số (Un) biết Un = 1/(1.3) + 1/(2.4) + 1/(3.5) +…+ 1/(n.(n+2))

Đáp án: $0<U_n<\dfrac34$

Giải thích các bước giải:

Ta có $U_n-u_{n-1}=\dfrac{1}{n\cdot (n+2)}>0$

$\to U_n>u_{n-1}$

$\to$Hàm số tăng

Ta có:

$U_n=\dfrac{1}{1\cdot 3}+\dfrac{1}{2\cdot 4}+\dfrac{1}{3\cdot 5}+…+\dfrac{1}{n\cdot (n+2)}$

$\to U_n=\dfrac12(\dfrac{2}{1\cdot 3}+\dfrac{2}{2\cdot 4}+\dfrac{2}{3\cdot 5}+…+\dfrac{2}{n\cdot (n+2)})$

$\to U_n=\dfrac12(\dfrac{3-1}{1\cdot 3}+\dfrac{4-2}{2\cdot 4}+\dfrac{5-3}{3\cdot 5}+…+\dfrac{n+2-n}{n\cdot (n+2)})$

$\to U_n=\dfrac12(\dfrac11-\dfrac13+\dfrac12-\dfrac14+\dfrac13-\dfrac15+…+\dfrac1n-\dfrac1{n+2})$

$\to U_n=\dfrac12((1+\dfrac12+\dfrac13+….+\dfrac1n)-(\dfrac13+\dfrac14+\dfrac15+…+\dfrac1{n+2})$

$\to U_n=\dfrac12(1+\dfrac12-\dfrac1{n+1}-\dfrac1{n+2})$

$\to U_n=\dfrac12(\dfrac32-\dfrac1{n+1}-\dfrac1{n+2})$

Mà $n\in N^*\to 0< n<+\infty$

$\to \dfrac12(\dfrac32-\dfrac1{0+1}-\dfrac1{0+2})< U_n <\dfrac12(\dfrac32-\dfrac1{+\infty+1}-\dfrac1{+\infty+2})$

$\to 0<U_n<\dfrac34$