Exercise 2.4.2

Question

\begin{enumerate}[label=(\alph*)] 
  \item Consider the recursively defined sequence $y_{1}=1$
  $$
  y_{n+1}=3-y_{n}
  $$
  and set $y=\lim y_{n} .$ Because $\left(y_{n}\right)$ and $\left(y_{n+1}\right)$ have the same limit, taking the limit across the recursive equation gives $y=3-y$. Solving for $y$, we conclude $\lim y_{n}=3 / 2$
  What is wrong with this argument?
  \item This time set $y_{1}=1$ and $y_{n+1}=3-\frac{1}{y_{n}}$. Can the strategy in (a) be applied to compute the limit of this sequence?
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item The sequence $y_n = (1, 2, 1, 2, \dots)$ does not converge.
  \item Yes, $y_n$ converges by the monotone convergence theorem since $0 < y_n < 3$ and $y_n$ is increasing.
   \end{enumerate}

Exercise 2.4.2

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Exercise 2.4.2

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