Let $ a_0 = 0 $ and define $ a_n = \frac{1}{2} (1 + a_{n-1}) $ for all positive integers $ n \geq 1 $. The least value of $ n $ for which $ |1 - a_n|

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Let $ a_0 = 0 $ and define $ a_n = \frac{1}{2} (1 + a_{n-1}) $ for all positive integers $ n \geq 1 $. The least value of $ n $ for which $ |1 - a_n|<\frac{1}{2^{10}} $ is ______.

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We are given the recurrence relation $ a_n = \frac{1}{2} (1 + a_{n-1}) $ with the initial condition $ a_0 = 0 $. We need to find the least value of $ n $ for which $ |1 - a_n| < \frac{1}{2^{10}} $. Step 1: Observing the recurrence relation, we see that the sequence $ a_n $ converges to 1. We can express $ a_n $ as: $$ a_n = 1 - \frac{1}{2^n}. $$ This is a standard result for such recurrence relations, where $ a_n $ converges to 1. Step 2: We need to find $ n $ such that: $$ |1 - a_n| = \left| 1 - \left( 1 - \frac{1}{2^n} \right) \right| = \frac{1}{2^n} < \frac{1}{2^{10}}. $$ This inequality simplifies to: $$ \frac{1}{2^n} < \frac{1}{2^{10}}. $$ Step 3: Solving for $ n $, we get: $$ n \geq 11. $$ Thus, the least value of $ n $ is 11 .

Let \( a_0 = 0 \) and define \( a_n = \frac{1}{2} (1 + a_{n-1}) \) for all positive integers \( n \geq 1 \). The least value of \( n \) for which \( |1 - a_n|<\frac{1}{2^{10}} \) is ______.

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