Giải thích các bước giải:
Ta có :
$u_n=\dfrac{n}{1+n^2+n^4}=\dfrac{n}{1+2n^2+n^4-n^2}=\dfrac{n}{(n^2+1)^2-n^2}=\dfrac{n}{(n^2+n+1)(n^2-n+1)}$
$\to u_n=\dfrac 12.\dfrac{n^2+n+1-(n^2-n+1)}{(n^2+n+1)(n^2-n+1)}=\dfrac{1}{2}(\dfrac{1}{n^2-n+1}-\dfrac{1}{n^2+n+1})$
$\to \begin{cases}u_1=\dfrac{1}{2}(\dfrac{1}{1^2-1+1}-\dfrac{1}{1^2+1+1})\\u_2=\dfrac{1}{2}(\dfrac{1}{2^2-2+1}-\dfrac{1}{2^2+2+1})\\...\\u_n=\dfrac{1}{2}(\dfrac{1}{n^2-n+1}-\dfrac{1}{n^2+n+1})\end{cases}$
$\to u_1+u_2+..+u_n=\dfrac{1}{2}(\dfrac{1}{1^2-1+1}-\dfrac{1}{n^2+n+1})$
$\to u_1+u_2+..+u_n=\dfrac{1}{2}(1-\dfrac{1}{n^2+n+1})$
$\to \lim (u_1+u_2+..+u_n)=\lim (\dfrac{1}{2}(1-\dfrac{1}{n^2+n+1}))=\dfrac{1}{2}\to A$
