Finding the limit of the sequence

Question

Let $a ∈ R$ Consider $x_{1} = a, x_{2} = (1+a)/2$ , and by induction $x_{n} := (1+x_{n−1})/2$ What is the limit $?$

By replacing $x_{n}$ and $x_{n-1}$ by $l$, we get the limit $l=1$. So limit should be $1$.

Also the nth term can be represented by $x_{n} = ( a+ 1 + 2 + 2^{2}+......+ 2^{n-2})/ 2^{n-1}$ And again the limit is $1$

I want to know, why does the value of $a$ doesn't affect the limit$?$ Does this sequence always converge$?$ And also what can I say about the monotonicity of the sequence$?$

Answer 1

As you've already determined,

$$\begin{equation}\begin{aligned} x_n & = \frac{a + 1 + 2 + 2^2 + \ldots + 2^{n-2}}{2^{n-1}} \\ & = \frac{a - 1}{2^{n-1}} + \frac{2 + 2 + 2^2 + \ldots + 2^{n-2}}{2^{n-1}} \\ & = \frac{a - 1}{2^{n-1}} + \frac{2^{n-1}}{2^{n-1}} \\ & = \frac{a - 1}{2^{n-1}} + 1 \end{aligned}\end{equation}\tag{1}\label{eq1}$$

Since $\frac{1}{2^{n-1}} \to 0$ as $n \to \infty$, the first term goes to $0$ regardless of what $a$ is, with the value of $a - 1$ only affecting how fast the first term goes to $0$. This also shows the sequence always converges to a limit of $1$.

As for the sequence's monotonicity, this depends on the value of $a - 1$. If it's $0$, the sequence is a constant of $1$, while if it's positive, then the sequence is monotonically decreasing, and if it's negative, then the sequence is monotonically increasing.