Evaluating the Convergence of a Sequence with Mathematical Induction

In this article, we will explore how to evaluate the convergence of a specific sequence using mathematical induction and limits. We will dive into the steps required to analyze the behavior of the sequence as it approaches infinity. This method is particularly useful when dealing with complex sequences that involve recursive definitions and limits.

Introduction

Consider the sequence defined by:

Definition of the Sequence

Let ( I lim_{n to infty} frac{a_n}{2^n} ) and ( a_{n 1} a_n sqrt{1 - a_n^2} ).

Step 1: Establishing the Monotonicity

To evaluate the sequence, we first need to determine if it is monotonically increasing. We start with the initial condition:

Let ( a_{n 1} a_n sqrt{1 - a_n^2} ).

From this definition, we know that ( a_{n 1} ) is always less than or equal to ( a_n ), but let us examine the inequality in detail:

Step 2: Establishing ( 2a_n leq a_{n 1} )

To prove that ( a_{n 1} geq 2a_n ), we start with the given condition:

Let ( b_n frac{a_n}{2^n} ).

Given ( a_{n 1} a_n sqrt{1 - a_n^2} ) implies:

2a_n leq a_{n 1}

Thus, we can rewrite the inequality as:

frac{2a_n}{2^{n 1}} leq frac{a_{n 1}}{2^{n 1}}

This implies that the sequence ( frac{a_{n 1}}{2^{n 1}} ) is monotonically increasing if ( 2a_n leq a_{n 1} ) and ( b_n frac{a_n}{2^n} ).

Step 3: Proving the Monotonicity by Induction

To prove the sequence ( frac{a_{n 1}}{2^{n 1}} ) is monotonically increasing, we use mathematical induction:

Base Case

For ( n 1 ), we assume ( a_{1 1} a_1 sqrt{1 - a_1^2} ).

Inductive Step

Assume ( frac{a_{k 1}}{2^{k 1}} leq frac{a_{k 2}}{2^{k 2}} ) is true for some ( k ). Prove that ( frac{a_{k 2}}{2^{k 2}} leq frac{a_{k 3}}{2^{k 3}} ).

This step involves showing that the inequality holds for all ( n ) by induction.

Step 4: Boundedness and Convergence

Given that the sequence is monotonically increasing, we need to determine if it is bounded above. If the sequence is bounded, it must converge to a limit:

Assuming Boundedness

Let ( b_n frac{a_n}{2^n} ). If the sequence ( b_n ) is bounded above, it converges to a limit:

Limit of the Sequence

Define ( lim_{n to infty} b_n b ).

Proof of Convergence

Since ( a_{n 1} a_n sqrt{1 - a_n^2} ), taking the limit:

lim_{n to infty} b_{n 1} lim_{n to infty} b_n

This implies:

b_{n 1}  b_n sqrt{1 - b_n^2}

Solving for ( b ) we find:

b b sqrt{1 - b^2}

This equation holds if ( b 0 ).

Conclusion

In conclusion, we have established that the sequence ( frac{a_n}{2^n} ) is monotonically increasing and bounded above, leading to its convergence. The limit of the sequence is ( b 0 ). This method of evaluation is crucial for understanding the behavior of recursive sequences and applying mathematical induction to solve complex problems.