Solutions

(a) We ought to first prove that a limit exists before assuming that it does. This argument assumes beforehand that the limit exists. In fact, it isn't hard to see that this sequence simply alternates between 1 and 2, causing it to diverge.

(b) It can, only if it converges. The burden now is on us to determine whether it does or doesn't.

On first inspection, the sequence seems to be increasing:

$$y_{n} = (1,\; 2,\; 2.5,\; 2.6,\; \dots).$$

This intuition should tempt us to try and apply the Monotone Convergence Theorem (MCT). Thus, we give in, and attempt to firstly prove that it is increasing. Using induction, the base case is easily proved:

$$y_2 = 2 > 1 = y_1.$$

Moving on to the induction step, we assume that $y_{n + 1} > y_{n}$ and obtain

$$\begin{aligned} y_{n + 2} - y_{n + 1} &= (3 - \frac{1}{y_{n + 1}}) - (3 - \frac{1}{y_{n}}) \\ &= \frac{1}{y_{n}} - \frac{1}{y_{n + 1}} \\ &= \frac{y_{n + 1} - y_{n}}{y_{n} y_{n + 1}} > 0. \end{aligned}$$

Now, we also prove that the sequence $y_{n}$ is bounded above. Intuitively, 3 seems like a fairly good candidate. Again, we use induction, to prove that it is a valid upper bound. The base case $y_1 = 1 < 3$ obviously holds true. For the induction step, assume that $y_{n} < 3$. We then have

$$\begin{aligned} y_{n + 1} = 3 - \frac{1}{y_{n}} < 3 &\iff \frac{1}{y_{n}} > 0 \\ &\iff y_{n} > 0. \end{aligned}$$

Focusing on the last inequality we need only prove that $y_{n} > 0$. Notice first that we proved that $y_{n}$ is increasing. Secondly, the lower bound of $y_{n}$ is 1 $(y_1 = 1)$. Therefore, we have proved that the sequence is bounded above and thus can be certain that the sequence possesses a limit.