Solutions

(a) This is possible. Consider the sequences

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

and

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

The sum of these two sequences gives us the constant sequence

$$x_{n} + y_{n} = (1, 1, 1, 1, \dots)$$

that clearly converges to 1, although both $x_{n}$ and $y_{n}$ diverge.

(b) Intuitively, this claim is false. Suppose it were true, to attempt to encounter a contradiction. That is, for any convergent sequence $x_{n}$ and divergent sequence $y_{n}$, their sum $c_{n} = x_{n} + y_{n}$ is a convergent sequence. Suppose further that $x_{n}$ converges to $l$ and that $c_{n}$ converges to $r$. Therefore, by the algebraic limit theorem,

$$\lim (c_{n} - x_{n}) = \lim c_{n} + \lim -x_{n} = r - l.$$

However,

$$\lim (c_{n} - x_{n}) = \lim(y_{n})$$

and we therefore have a contradiction since $y_{n}$ was not supposed to converge.

(c) Consider the convergent sequence

$$b_{n} = (1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots)$$

respecting the condition that

$$\forall n \in \mathbb{N}, \quad b_{n} \neq 0.$$

Clearly, the sequence

$$\frac{1}{b_{n}} = (1, 2, 4, 8, \dots)$$

diverges.

(d) Difficult :(

(e) Take any possible divergent sequence $b_{n}$ imaginable. Now consider the sequence

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

which clearly converges to 0. If we multiply both sequences together, we end up with the sequence

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

which also converges.