Solutions

(a) False. Consider the sequences

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

Clearly the sequence $( a_{n} - b_{n} )$ converges to 0 (why). However, neither of the sequences converge.

(b) True. By definition, for every $\epsilon > 0$,

$$\exists N \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N \implies \abs{b_{n} - b} < \epsilon.$$

However, by using the reverse triangle inequality, we also have

$$n \geq N \implies \abs{\abs{b_{n}} - \abs{b}} \leq \abs{b_{n} - b} < \epsilon.$$

(c) True. Suppose

$$a_{n} \rightarrow 0, \quad c_{n} = (b_{n} - a_{n}) \rightarrow 0.$$

Apply rule (ii) of the algebraic limit theorem to the limit

$$\lim (a_{n} + c_{n}) = \lim a_{n} + \lim c_{n} = 0 + 0 = 0.$$

This then implies that $\lim b_{n} = 0$ since

$$\lim (a_{n} + c_{n}) = \lim a_{n} + b_{n} - a_{n} = \lim b_{n}.$$

(d) True. Since

$$\forall n \in \mathbb{N}, \quad \abs{b_{n} - b} \leq a_{n},$$

every term of the sequence $a_{n}$ is positive. Therefore, by definition, for any $\epsilon > 0$,

$$\exists N \in \mathbb{N}, \forall n \in \mathbb{N}, \quad n \geq N \implies \abs{b_{n} - b} \leq a_{n} = \abs{a_{n}} < \epsilon.$$

Therefore, $b_{n}$ converges to $b$.