Finding the limit of the sequence 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$?$

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.
