Solutions

(a) We are not allowed to use the algebraic limit theorem because there is no guarantee that $a_{n}$ converges to a given value.

For the actual proof, suppose $a_{n}$ is bounded by $M > 0$ and that $b_{n}$ converges to 0. By definition, for every $\epsilon > 0$, we have

$$\exists N \in \mathbb{N}, \quad n \geq N \implies \abs{a_{n}b_{n}} \leq \abs{a_{n}M} < \frac{\epsilon}{M} M = \epsilon$$

(b) We do have an example of a sequence $a_{n}b_{n}$ which does not converge. Suppose

$$a_{n} = (0, 1, 0, 1, 0, \dots)$$

and

$$b_{n} = (1, 1, 1, 1, \dots).$$

Obviously $a_{n}$ is bounded by 1 and $b_{n} \rightarrow 1 \neq 0$. However, the sequence $a_{n}b_{n}$ simply is

$$a_{n}b_{n} = a_{n} = a_{n} = (0, 1, 0, 1, 0, \dots)$$

so $a_{n}b_{n}$ diverges in this case. On the other hand, if $a_{n}$ were equal to $b_{n}$, then our sequence would be convergent (why). Therefore, it is not possible to establish whether the sequence $a_{n}b_{n}$ converges or not.

(c) Suppose any sequence $a_{n}$ and $b_{n}$ such that

$$\begin{aligned} a_{n} \rightarrow 0 \\ b_{n} \rightarrow b. \end{aligned}$$

Since $b_{n}$ is convergent, it also is bounded then. Therefore we can apply the proposition from (a) and obtain

$$\lim a_{n} b_{n} = 0,$$

thereby proving rule (iii) from the algebraic limit theorem when $a_{n} \rightarrow 0$.