Does this sequence diverge to ∞?

Question

The sequence $(a_n)_{n \geq 1}$ is defined as follows:

$$a_n:= \begin{cases} 0 \quad \text{if} \quad n \quad \text{is odd}\\ n \quad \text{if} \quad n \quad \text{is even}\end{cases} \quad .$$

Does $(a_n)_{n \geq 1}$ diverge to $+\infty \quad ?$

Answer 1

The sequence is actually oscillating sequence. You can't say this sequence diverge to ∞. Because of the definition of divergence to infinity. But it has a divergent subsequence.

Answer 2

no, because a sequence is divergent to $+\infty$ if and only if (by definition): $$ \forall M\in\mathbb{R^+}:\exists\nu\in\mathbb{N}\ \ |\ \ n\gt\nu \implies a_n > M $$ which is obviously false since when n is odd: $$ \forall M\in\mathbb{R^+},n\in\mathbb{N},n \mbox{ odd}:\ \ \ a_n < M \ \ \ \ \ (\mbox{because } a_n=0) $$

Answer 3

A sequence $a_n$ diverges to infinity if and only if every subsequence of $a_n$ is unbounded above.

Using this information, you tell me: is the even subsequence $a_{2k}$ unbounded above? What about the odd subsequence, $a_{2k+1}$? What can you conclude from this?