91Ó°ÊÓ

The arithmetic mean is the simple average of two numbers. In our sequence, each term is formed by taking the average of the number \(\alpha\) and the previous term. This is represented by the formula \(a_n = \frac{\alpha + a_{n-1}}{2}\). By continuously averaging \(\alpha\) with preceding terms, each new value of the sequence \(\{a_n\}\) slightly adjusts in proximity to \(\alpha\).
This averaging process is crucial because it means each step systematically moves the sequence closer to the desired convergence point. Over many iterations, the arithmetic mean smoothens the trajectory of the sequence towards being stable at \(\alpha\).

A recursive sequence is one where each term is defined as a function of one or more of the preceding terms. In this problem, the sequence begins with a starting term, \(a_1 = 1\), followed by the definition \(a_n = \frac{\alpha + a_{n-1}}{2}\) for \(n > 1\).
This implies that to find any term in the sequence, you need to know the term before it, thus establishing a chain reaction or a recursive nature. The beauty of recursive sequences lies in their pattern. By frequently combining with \(\alpha\), it ensures systematic progression towards a specific limit. This step-by-step refinement guides us towards convergence, demonstrating how recursive sequences are powerful tools for establishing mathematical limits.

Convergence is about whether a sequence approaches a specific value as it progresses to infinity. In this context, we're interested in if \(\{a_n\}\) converges to \(\alpha\). The observation is that, with each new term being closer to \(\alpha\), we conjecture that indeed the entire sequence will converge to this value in the limit.
Our hypothesis is based on using the sequence \(b_n = \alpha - a_n\), which simplifies the convergence analysis. Since \(b_n = \frac{b_{n-1}}{2}\), this halves itself with every step, demonstrating geometric contraction. When terms contract geometrically, they eventually converge to zero. Thus, the sequence \(\{a_n\}\) converges to \(\alpha\), confirming our hypothesis.

Although our problem involves convergence, exponential growth plays a role in proving it. Specifically, consider the term \( |\alpha - a_n| = \frac{|\alpha - 1|}{2^{n-1}} \).
The denominator \(2^{n-1}\) increases exponentially as \(n\) becomes large, making \( |\alpha - a_n|\) smaller rapidly. This demonstrates how the gap between \(a_n\) and \(\alpha\) decreases sharply. The rapid growth of the exponential term ensures that any initial difference \(\alpha - a_1\) becomes negligible over time.

• An exponential growth factor ensures fast convergence.
• The rate of shrinkage in \(b_n\) is dramatic due to this exponential property.