Step 1 :Rewrite the function $F(x)=x^{2}+2 x+1$ in vertex form as $F(x)=(x+1)^2$
Step 2 :Identify the vertex of the parabola as $(-1,0)$
Step 3 :Since the coefficient of $x^2$ is positive, the parabola opens upwards, meaning the vertex is the minimum point of the parabola
Step 4 :Therefore, the range of the function is $y\geq0$ or in interval notation, $[0, \infty)$
Step 5 :\(\boxed{[0, \infty)}\) is the final answer