{1+1/ n ∨ n∈N}
x1=1+1/1=2
x2=1+1/2=1.5
1.5-2=-0.5
because of x2<x1 the sequence monotonously decreases
Since for all n,xn>0, "\\lim _{n\\to \\infty }\\left(1+\\frac{1}{n}\\right)=1" then it is bounded below
{−1−1/ n ∨ n∈N}
x1=-1-1=-2
x2=-1-1/2=-1.5
-1.5-(-2)=0.5
because of x2>x1 the sequence monotonously increases
Since for all n,xn>0 "\\lim _{n\\to \\infty }\\left(-1-\\frac{1}{n}\\right)=-1" then it is bounded above
which means that the limit of the set exists.
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