Answer
$\displaystyle\lim_{n\to\infty} a_n = 1$
Work Step by Step
Step 1. Framing the Conjecture.
The terms of the sequence $\{a_{n+1}\}_{n=1}^\infty$ are obtained by removing the first term of the sequence $\{a_n\}_{n=1}^\infty$ without disturbing the order of the rest of the terms. So, the two sequences have the same terms except for the first term. In this sense, the sequence $\{a_{n+1}\}_{n=1}^\infty$ is a subsequence of the sequence $\{a_n\}_{n=1}^\infty$. The fact that the sequence $\{a_n\}_{n=1}^\infty$ converges to some number $L$ means that $a_n$ approaches $L$ as $n$ becomes large. The sequence $\{a_{n+1}\}_{n=1}^\infty$ has the same terms as $\{a_n\}_{n=1}^\infty$ except for the first term and in the same order. The recurrence relation $a_{n+1} = \frac{1}{2}(a_n + 1)$ would imply that $L = \frac{1}{2}(L + 1)$, and so $L$ would have the value $1$.
Conjecture: \[ \lim_{n\to\infty} a_n = 1 \]
Step 2. Proving the Conjecture.
We repeatedly use the recurrence relation $a_{n+1} = \frac{1}{2}(a_n + 1)$ to obtain: \[ \begin{align*} a_2 &= \frac{1}{2}(a_1 + 1) \\ a_3 &= \frac{1}{2}(a_2 + 1) = \frac{1}{2}\left(\frac{1}{2}(a_1 + 1) + 1\right) \\ a_4 &= \frac{1}{2}(a_3 + 1) = \frac{1}{2}\left(\frac{1}{2}\left(\frac{1}{2}(a_1 + 1) + 1\right) + 1\right) \\ &\vdots \end{align*} \] In general: \[ a_n = \frac{1}{2^{n-1}}a_1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots + \frac{1}{2^{n-1}} \]
Taking the limit as $n$ tends to $\infty$: \[ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{1}{2^{n-1}}a_1 + \lim_{n\to\infty} \left(\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots + \frac{1}{2^{n-1}}\right) \] The series $\sum_{n=1}^\infty \left(\frac{1}{2}\right)^n$ is a geometric series with $a = \frac{1}{2}$ and $r = \frac{1}{2}$, and so its value is: \[ \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n = \frac{a}{1-r} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1 \]
Therefore: \[ \lim_{n\to\infty} a_n = 0\cdot a_1 + 1 = 1 \] This proves the conjecture. Result \[ \lim_{n\to\infty} a_n = 1 \]
