Ta có: $x_{n+1}-x_{n}=\frac{(2x_{n}+1)^2}{2}(1) \geq 0 \Rightarrow$ dãy tăng.
Giả sử dãy số bị chặn trên. Gọi $lim x_{n}=a\geq 1 \Rightarrow a= \frac{(2a+1)^2}{2}+a\Rightarrow a=\frac{-1}{2}$( loại)
$\Rightarrow lim x_{n}=+\propto$
Từ (1) $\Rightarrow \frac{2x_{n}+1}{2x_{n+1}+1}= \frac{2(x_{n+1}-x_{n})}{(2x_{n}+1)(2x_{n+1}+1)}= \frac{1}{2x_{n}+1}-\frac{1}{2x_{n+1}+1}$
$\Rightarrow \lim\sum_{i=1}^{n}\frac{2x_{i}+1}{2x_{i+1}+1}=lim(\frac{1}{2x_{1}+1}-\frac{1}{2x_{n+1}+1})=\frac{1}{3}$