The graph shows the sequence does not converges only at x=1 because we set x(1) = 1. Elsewhere the graph does not converge. The limit of \(x(n)\) is from 1 to \(\infty\) as \(n\geq 1\) goes to \(\infty\).
The graph shows the sequence does not converge only at x(n)=1 because we set x(1) = 1. Elsewhere the graph does not converge. The limit of \(x(n)\) is from 1 to \(\infty\) as \(n\geq 1\) goes to \(\infty\).
The sequence x(n) does converge at an estimated value of \(x(n) = 1\frac{2}{3}\). I think this because as n increases to \(\infty\) the values of x(n) get closer to the value \(1\frac{2}{3}\). If we look at the graph it becomes more horizontal close to a straight line at \(x(n) = 1\frac{2}{3}\).
\[\vert x(n+1)-x(n)\vert = \frac{1}{2^{n-1}}\]
\(x = 0\) \(\rightarrow\) \(\vert x_{2} - x_{1} \vert\) \(\rightarrow\) \(\vert 2-1 \vert\) \(\rightarrow\) 1 = \(\frac{1}{2^0} \)
\(x = 1\) \(\rightarrow\) \(\vert x_{3} - x_{2} \vert\) \(\rightarrow\) \(\frac{1}{2}\vert x_{2} - x_{1} \vert\) \(\rightarrow\) \(\frac{1}{2}\) = \(\frac{1}{2^1}\)
\(x = 2\) \(\rightarrow\) \(\vert x_{4} - x_{3} \vert\) \(\rightarrow\) \( \frac{1}{2} \vert x_{3} - x_{2} \vert\) \(\rightarrow\) \(\frac{1}{4}\) = \(\frac{1}{2^2}\)
\(x = 3\) \(\rightarrow\) \(\vert x_{5} - x_{4} \vert\) \(\rightarrow\) \( \frac{1}{2} \vert x_{4} - x_{3} \vert\) \(\rightarrow\) \(\frac{1}{8}\) = \(\frac{1}{2^3}\)
Cauchy sequence
[x_1, x_2 , x_3, x_4 , ...x_i]
\(\vert x_{n+p} - x_{n} \vert\) for \(p \geq 1\)
x_n is a Cauchy sequence if for all \(\epsilon \geq 0\) there is an N such that. \[x_{n+p} - x_{n} < \epsilon\] where \(n \geq N\) and \(p \geq 1\)
\[\vert x(n+1)-x(n)\vert = \frac{1}{2^{n-1}}\]
\[\frac{1}{2^{n+p-2}} + \frac{1}{2^{n+p-3}} + \dots + \frac{1}{2^{n-1}}\] \[\frac{1}{2^{n-1}} (\frac{1}{2^{p-1}} + \frac{1}{2^{p-2}} + \dots + 1)\]