real analysis - Proving that Cauchy condensation test is true for $a_n$ any monotone sequence.

I'm trying to show that then $\Sigma a_n$ converges iff $\Sigma 2^n a_{2^n}$ converges if $a_n$ is any monotone sequence.

I am assuming I have to prove the cases:

• $a_n$ is positive and increasing


• $a_n$ is negative and increasing


• $a_n$ is positive and decreasing


• $a_n$ is negative and decreasing

Here is what I have worked out from some of the comments below:

Assume $a_n$ is an increasing sequence. If $a_n$ is positive, then $\lim a_n \neq 0$ and we have divergence.

I don't really understand why this is. Is it because $a_n$ is unbounded and so it diverges to $\infty$? I don't know a rigorous way to prove this. I'm assuming I have to make use of the theorem that monotonic $s_n$ converges iff it is bounded.

If $a_n$ is negative, then we have that $\lim a_n = 0$

Again, I don't know how to rigorously prove this. How do I find a bound for $a_n$?

Let $s_n = a_1 + a_2 + ... + a_n$ and $t_n = a_1 + 2 a_2 + ... + 2^k a_{2^k}$.
For $n<2^k$, we have $s_n \geq a_1 + (a_2 + a_3) + ... _ (a_{2^k}+...+a_{2^{k+1}-1}
\geq a_1 + a_2 + ... + 2^{k}a_{2k} = t_k
$

And so $s_n \geq t_k$.

Similar proof for $2s_{n} \leq t_k$

Someone below has already proved nonnegative decreasing case. I'm assuming that:

If $a_n$ was decreasing and negative, then $\lim a_n \neq 0$ and we have divergence.

Again, I'm not sure how to prove this.