11.1 Subsequences

Workshop 3.pdf

1. (a) Consider the sequence \(y_1 = 1\) and \(y_{n+1} = 3 - y_n\), and set \(y = \lim y_n\). Since \((y_n)\) and \((y_{n+1})\) have the same limit, taking the limit across the recursive equation gives \(y = 3 - y\). Solving, we conclude that \(\lim y_n = \frac{3}{2}\). What is wrong with this argument?

   No, because limit doesn't exist

   (b) This time, set \(y_1 = 1\) and \(y_{n+1} = 3 - \frac{1}{y_n}\). Find the limit of the sequence.

   Since \(y_n\) has limits, then \(\lim_{n \to \infty} y_n = \lim_{n \to \infty} y_{n+1} = y\).

   Thus, \(y = 3 - \frac{1}{y} \Rightarrow y^2 - 3y + 1 = 0 \Rightarrow y = \frac{3 \pm \sqrt{5}}{2}\).

   To determine which solution is correct, we use induction to show that \(1 \leq y_n \leq 3\) for all \(n \geq 1\).

   Base Case: For \(n = 1\), \(1\leq y_1=1\leq y_2=2\leq3\).

   Inductive Step: Assume \(1\leq y_{n}\leq y_{n+1}\leq3\).

   We need to prove \(1\leq y_{n+1}\leq y_{n+2}\leq3\).

   Assume \(1\leq y_{n}\leq y_{n+1}\leq3\Rightarrow\frac{1}{y_{n}}\geq\frac{1}{y_{n+1}}\Rightarrow3-\frac{1}{y_{n}}\leq3-\frac{1}{y_{n+1}}\Rightarrow y_{n+1}\leq y_{n+2}\).

   Then it is monotonic increasing.

   Since \(y_{n+2}=3-\frac{1}{y_{n+1}}\leq3\), then we have proved \(\forall n,1\leq y_n\leq 3\)

   Therefore, \(\lim_{n \to \infty} y_n = \frac{3 + \sqrt{5}}{2}\).

2. (a) Prove that the sequence defined by \(x_1 = 3\) and \(x_{n+1} = \frac{1}{4 - x_n}\) converges.

   (b) Compute \(\lim x_n\).

   Easy

3. Decide whether the following propositions are true or false, providing a short justification for each conclusion.

   (a) If every proper subsequence of \((x_n)\) converges, then \((x_n)\) converges as well.

   (b) If \((x_n)\) contains a divergent subsequence, then \((x_n)\) diverges.

   (c) If \((x_n)\) is monotone and contains a convergent subsequence, then \((x_n)\) converges.

‍