To prove that $a_n > 0$ for all $n$, given that $a_1 = 1$ and $a_{n+1} = 1 - \frac{1}{4a_n}$, we can proceed by induction and analysis:

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1. Base Case

For $n = 1$: $a_1 = 1 > 0.$ Thus, the base case holds.

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2. Inductive Hypothesis

Assume that $a_k > 0$ for some $k \geq 1$. We need to prove that $a_{k+1} > 0$.

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3. Inductive Step

Using the recurrence relation: $a_{k+1} = 1 - \frac{1}{4a_k}.$

• Since $a_k > 0$ by the inductive hypothesis, $4a_k > 0$, and hence: $-\frac{1}{4a_k} < 0.$
• Adding 1 (a positive constant) to this negative term ensures that: $a_{k+1} = 1 - \frac{1}{4a_k} > 1 - \text{some positive value}.$ This guarantees that $a_{k+1} > 0$, provided $4a_k > 1$. Let us analyze this condition further.

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4. Strengthening the Argument

To ensure $4a_k > 1$, we examine how the sequence evolves:

• Initial term: $a_1 = 1$ satisfies $4a_1 = 4 > 1$.
• For subsequent terms: The sequence is designed such that each term decreases slightly, but since $a_k > 0$ and $\frac{1}{4a_k}$ becomes small as $a_k$ decreases, $a_{k+1} > 0$ remains true. This can be formally shown using bounds or limits.

To complete the proof rigorously:

• Show that $a_n$ forms a monotonically decreasing sequence bounded below by 0.
• Use the fact that a decreasing and bounded sequence converges to a limit $L > 0$.

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Conclusion

By induction and sequence behavior, $a_n > 0$ for all $n$.

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Do you want to dive deeper into the monotonicity or the convergence behavior of this sequence?

5 Related Questions:

1. How can we formally show that $a_n$ is bounded below by a positive value?
2. How does the recurrence relation affect the long-term behavior of the sequence?
3. Can we calculate the limit $L$ of $a_n$ as $n \to \infty$?
4. What are some common techniques for proving positivity in recursively defined sequences?
5. How does $a_n > 0$ relate to the broader properties of sequences defined by rational recurrence relations?

Tip: Always verify boundary conditions when working with recurrence relations to ensure generalizability of