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 know how to prove Cauchy condensation when $a_n$ is decreasing and nonnegative, but do I also have to consider other cases such as when $a_n$ is decreasing and negative, $a_n$ is increasing and positive and $a_n$ is increasing and negative?

I feel like I'm approaching this problem in the wrong way. Help would be much appreciated!

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.

I know that monotone sequences can be increasing or decreasing.

For the decreasing case, $\Sigma a_n$ converges iff $s_n$ converges iff $s_n$ is bounded. So I need to show that somehow $S_{2^n}$ is also bounded? But then since this is any monotone sequence, do I need to show two other cases for nonnegative and negative sequences? What if the sequence contains both negative and nonnegative values?

I feel like I'm approaching this problem in the wrong way. Help would be much appreciated!

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 know how to prove Cauchy condensation when $a_n$ is decreasing and nonnegative, but do I also have to consider other cases such as when $a_n$ is decreasing and negative, $a_n$ is increasing and positive and $a_n$ is increasing and negative?

I feel like I'm approaching this problem in the wrong way. Help would be much appreciated!

I know that monotone sequences can be increasing or decreasing.

For the decreasing case, $\Sigma a_n$ converges iff $s_n$ converges iff $s_n$ is bounded. So I need to show that somehow $S_{2^n}$ is also bounded? But then since this is any monotone sequence, do I need to show two other cases for nonnegative and negative sequences? What if the sequence contains both negative and nonnegative?

I feel like I'm approaching this problem in the wrong way. Help would be much appreciated!

I know that monotone sequences can be increasing or decreasing.

For the decreasing case, $\Sigma a_n$ converges iff $s_n$ converges iff $s_n$ is bounded. So I need to show that somehow $S_{2^n}$ is also bounded? But then since this is any monotone sequence, do I need to show two other cases for nonnegative and negative sequences? What if the sequence contains both negative and nonnegative values?

I feel like I'm approaching this problem in the wrong way. Help would be much appreciated!