Question

Suppose that \(a_{1}=S_{1}=1\) and that, for given numbers \(S>1\) and \(0

Short answer

Yes, \( S_n \) converges to \( S \).

Step-by-step solution

Verify Inequality

Let's consider the sequence given by \( S_n \) defined as \( S_n = a_n + S_{n-1} \). Since initially \( S_1 = 1 \), and \( a_{n+1} = k(S - S_n) \) with \( 0 < k < 1 \), it follows that \( a_{n+1} = k(S - S_n) < S - S_n \). Therefore, \( S_{n+1} = a_{n+1} + S_n < S_n + (S - S_n) = S \). Thus, \( S_{n} < S \) for all iterations.

Establish Monotonicity

We use the definition of \( a_{n+1} \) to show that \( S_n \) is increasing. Observe that \( S_{n+1} = a_{n+1} + S_n = k(S - S_n) + S_n = S_n + k(S - S_n) \). Since \( 0 < k < 1 \) and \( S_n < S \), this implies \( S - S_n > 0 \) and \( k(S - S_n) > 0 \). Therefore, \( S_{n+1} = S_n + k(S - S_n) > S_n \), showing that \( S_n \) is indeed increasing.

Apply the Convergence Theorem

Since \( S_n \) is increasing and bounded above by \( S \), we can apply the Monotone Convergence Theorem. This theorem tells us that a sequence that is both increasing and bounded will converge to some limit \( L \), where \( L \leq S \). Thus, \( S_n \) has a limit.

Determine the Limit

To find the limit \( L \), assume \( S_n \to L \) as \( n \to \infty \). From the recurrence relation, we have \( a_{n+1} \to k(S - L) \), leading to \( S_{n+1} = a_{n+1} + S_n \to L = k(S - L) + L \). Solving \( k(S - L) = 0 \), we find \( S - L = 0 \) since \( k > 0 \). Hence, \( L = S \).

Key concepts

Monotone Convergence Theorem

The Monotone Convergence Theorem is a key concept in mathematical analysis that helps us determine whether a sequence converges. This theorem states that if a sequence is both monotonic (either entirely non-decreasing or non-increasing) and bounded, then it converges to a limit.
In our exercise, the sequence we are evaluating is \( S_n \). We find that \( S_n \) is an increasing sequence and has an upper bound, \( S \).
- **Monotonic:** This means that the sequence either always increases or always decreases. In our exercise, since \( S_n \leq S \), but always grows larger than any previous term, we verify the sequence is monotonic.- **Bounded:** Despite being ever increasing, the sequence never surpasses the upper bound \( S \). This containment ensures the sequence doesn't 'grow without control', which wouldn't allow convergence.
The theorem ensures that under these conditions, \( S_n \) converges. Therefore, in our specific example, because \( S_n \) is increasing and bounded, it must reach some limit.

Increasing Sequences

An increasing sequence is one where each term is larger or equal to the one before it. In simple terms, every step you take doesn't take you back or leave you in the same place; instead, it brings you forward.
In the sequence \( S_n \), using the formula \( S_{n+1} = S_n + k(S - S_n) \), we can show that \( S_{n+1} > S_n \) because:

• The expression \( k(S - S_n) \) is positive. This occurs because \( S \) is greater than any \( S_n \) we've obtained so far, keeping \( S - S_n \) positive.
• Multiplying by the positive \( k \) (since \( 0 < k < 1 \)) ensures that the sequence is growing.

Putting it all together, the new term \( S_{n+1} \) will always be larger than the previous term \( S_n \), confirming it's an increasing sequence.

Bounded Sequences

A bounded sequence is one that stays within some fixed limits, not straying off too far in any direction. In mathematical terms, a sequence \( S_n \) being bounded means there exist numbers \( M \) and \( m \) such that every term in the sequence satisfies \( m \leq S_n \leq M \).
- **Upper Bound:** Our sequence \( S_n \) is bounded above by \( S \). This is proved since every iteration keeps \( S_n \) less than \( S \), rewriting the familiar inequality, \( S - S_n > 0 \), addressed in the step-by-step solution. - **Lower Bound:** From the start, we are given \( S_1 = 1 \), which acts as a lower limit. This prevents our sequence from dropping lower unexpectedly.
Combined, these boundaries ensure that \( S_n \) doesn't veer off course, aiding in proving its convergence through the Monotone Convergence Theorem.