Ta có: [tex]U_{n+1}=\frac{1}{2}U_n-U_n+2\Rightarrow U_{n+1}-U_n=\frac{1}{2}U_n^2-2U_n+2=\frac{1}{2}(U_n-2)^2[/tex]
Lại có : [tex]U_{n+1}=\frac{1}{2}U_n^2-U_n+2\Leftrightarrow 2U_{n+1}=U_n^2-2U_n+4\Leftrightarrow 2(U_{n+2}-2)=U_n(U_n-2)\\\Rightarrow \frac{1}{U_n}=\frac{1}{2}.\frac{U_n-2}{U_{n+1}-2}=\frac{1}{2}\frac{(U_n-2)^2}{(U_n-2)(U_{n+1}-2)}=\frac{U_{n+1}-U_n}{(U_n-2)(U_{n+1}-2)}\\\Leftrightarrow \frac{1}{U_n}=\frac{1}{U_n-2}-\frac{1}{U_{n+1}-2}[/tex]
Do đó : [tex]V_n=\sum \frac{1}{U_k}=\frac{1}{U_1-2}-\frac{1}{U_{n+1}-2}=2-\frac{1}{U_{n+1}-2}[/tex]
Tiếp theo chứng minh [tex]\lim_{n\rightarrow +\infty }U_n=+\infty[/tex] suy ra $LimV_n=2$