Xét tính bị chặn của dãy số (Un) biết Un = 1/(1.3) + 1/(2.4) + 1/(3.5) +…+ 1/(n.(n+2)) 16/11/2021 Bởi Melanie 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 $n\in N^*\to 0< n<+\infty$ $\to \dfrac12(\dfrac32-\dfrac1{0+1}-\dfrac1{0+2})< U_n <\dfrac12(\dfrac32-\dfrac1{+\infty+1}-\dfrac1{+\infty+2})$ $\to 0<U_n<\dfrac34$ Bình luận
$u_n=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+…+\dfrac{1}{n(n+2)}$ $2u_n=1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+…+\dfrac{1}{n}-\dfrac{1}{n+2}$ $=1+\dfrac{1}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}$ $=\dfrac{3}{2}+\dfrac{-n-2-n-1}{(n+1)(n+2)}$ $=\dfrac{3}{2}+\dfrac{-2n-3}{(n+1)(n+2)}$ $=\dfrac{3(n^2+3n+2)-2n-3}{2(n+1)(n+2)}$ $=\dfrac{3n^2+7n+3}{2(n+1)(n+2)}$ $\Leftrightarrow u_n=\dfrac{3n^2+7n+3}{4(n+1)(n+2)}$ $=\dfrac{3n^2+7n+3}{4n^2+12n+8}>0$ $\Rightarrow (u_n)$ bị chặn dưới $\lim u_n=\lim\dfrac{3n^2+7n+3}{4n^2+12n+8}$ $=\lim\dfrac{3+\dfrac{7}{n}+\dfrac{3}{n^2}}{4+\dfrac{12}{n}+\dfrac{8}{n^2}}$ $=\dfrac{3}{4}$ $\Rightarrow u_n<\dfrac{3}{4}$ $\Rightarrow (u_n)$ bị chặn trên Vậy $(u_n)$ là dãy bị chặn Bình luận
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 $n\in N^*\to 0< n<+\infty$
$\to \dfrac12(\dfrac32-\dfrac1{0+1}-\dfrac1{0+2})< U_n <\dfrac12(\dfrac32-\dfrac1{+\infty+1}-\dfrac1{+\infty+2})$
$\to 0<U_n<\dfrac34$
$u_n=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+…+\dfrac{1}{n(n+2)}$
$2u_n=1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+…+\dfrac{1}{n}-\dfrac{1}{n+2}$
$=1+\dfrac{1}{2}-\dfrac{1}{n+1}-\dfrac{1}{n+2}$
$=\dfrac{3}{2}+\dfrac{-n-2-n-1}{(n+1)(n+2)}$
$=\dfrac{3}{2}+\dfrac{-2n-3}{(n+1)(n+2)}$
$=\dfrac{3(n^2+3n+2)-2n-3}{2(n+1)(n+2)}$
$=\dfrac{3n^2+7n+3}{2(n+1)(n+2)}$
$\Leftrightarrow u_n=\dfrac{3n^2+7n+3}{4(n+1)(n+2)}$
$=\dfrac{3n^2+7n+3}{4n^2+12n+8}>0$
$\Rightarrow (u_n)$ bị chặn dưới
$\lim u_n=\lim\dfrac{3n^2+7n+3}{4n^2+12n+8}$
$=\lim\dfrac{3+\dfrac{7}{n}+\dfrac{3}{n^2}}{4+\dfrac{12}{n}+\dfrac{8}{n^2}}$
$=\dfrac{3}{4}$
$\Rightarrow u_n<\dfrac{3}{4}$
$\Rightarrow (u_n)$ bị chặn trên
Vậy $(u_n)$ là dãy bị chặn