Giúp e vs ạ Lim n!/(1+1^2)×(1+2^2)…(1+n^2) bằng ? 21/10/2021 Bởi Hailey Giúp e vs ạ Lim n!/(1+1^2)×(1+2^2)…(1+n^2) bằng ?
Đáp án: $0$ Giải thích các bước giải: Đặt $ $u_n=\dfrac{n!}{(1+1^2)(1+2^2)…(1+n^2)}$ $\to u_{n+1}=\dfrac{(n+1)!}{(1+1^2)(1+2^2)…(1+(n+1)^2)}$ $\to u_{n+1}=\dfrac{n!(n+1)}{(1+1^2)(1+2^2)…(1+n^2)(1+(n+1)^2)}$ $\to u_{n+1}=u_n\cdot \dfrac{n+1}{1+(n+1)^2}$ Đặt $\lim u_n=\lim u_{n+1}=a$ $\to \lim u_{n+1}=\lim u_n\cdot \dfrac{n+1}{1+(n+1)^2}$ $\to a=a\cdot \lim \dfrac{n+1}{1+(n+1)^2}$ $\to a=a\cdot \lim \dfrac{\dfrac1n+\dfrac1{n^2}}{\dfrac1{n^2}+(1+\dfrac1n)^2}$ $\to a=a\cdot 0$ $\to a=0$ Bình luận
Đáp án: $0$
Giải thích các bước giải:
Đặt $
$u_n=\dfrac{n!}{(1+1^2)(1+2^2)…(1+n^2)}$
$\to u_{n+1}=\dfrac{(n+1)!}{(1+1^2)(1+2^2)…(1+(n+1)^2)}$
$\to u_{n+1}=\dfrac{n!(n+1)}{(1+1^2)(1+2^2)…(1+n^2)(1+(n+1)^2)}$
$\to u_{n+1}=u_n\cdot \dfrac{n+1}{1+(n+1)^2}$
Đặt $\lim u_n=\lim u_{n+1}=a$
$\to \lim u_{n+1}=\lim u_n\cdot \dfrac{n+1}{1+(n+1)^2}$
$\to a=a\cdot \lim \dfrac{n+1}{1+(n+1)^2}$
$\to a=a\cdot \lim \dfrac{\dfrac1n+\dfrac1{n^2}}{\dfrac1{n^2}+(1+\dfrac1n)^2}$
$\to a=a\cdot 0$
$\to a=0$