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 $\lim u_n=\lim u_{n+1}=a$
$\to \lim u_{n+1}=\lim u_n\cdot \dfrac{n+1}{1+(n+1)^2}$
$\to a=a\cdot \lim \dfrac{n+1}{1+(n+1)^2}$
$\to a=a\cdot \lim \dfrac{\dfrac1n+\dfrac1{n^2}}{\dfrac1{n^2}+(1+\dfrac1n)^2}$
$\to a=a\cdot 0$
$\to a=0$
