There was an excercise About condensed sums. We should prove that:

If $ (a_n)$ is **monotonous** then $ \sum_{n}a_n$ converges/diverges iff $ \sum_{k}2^ka_{2^k}$ converges/diverges

It is sufficient to look at a decreasing Zero convergent sequence $ (a_n)$

$ s_n=\sum_{k=1}^{n}a_k,v_n=\sum_{k=1}^n2^ka_{2^k}$

If $ K,N\in\mathbb{N}$ with $ 2^{K-1}<N\leq 2^K$ then

I am sorry for the duplicate please answer one specific Question About the Problem I have in the comment and I will delete the answer afterwards

$ S_N\le S_{2^K}\leq a_1+v_{K-1}$ and

$ S_N\ge S_{2^{K-1}}\geq a_1/2 +v_{k-1}/2$

I have understood that the partialsums are Always positive since every $ a_n$ must be positive. And if the partialsums can be bounded then because the partialsums are monotonously rising the sequence of partialsums is converging to its Supremum. And $ s_k$ is bounded if and only if $ v_k$ is bounded.

Again because $ a_n$ is decreasing and zeroconvergent the partialsums are always positive. If one cannot be boundet then also the other one cannot be bounded.

**My Question** is that in the excercise the Task was to Show that the Statement holds if $ (a_n)$ is monotonous. But we have only looked at decreasing monotonous $ (a_n)$ and what is more important we have assumed that $ (a_n)$ is Zero convergent.

What if $ (a_n)$ is not Zero convergent then $ \sum_n a_n$ necessarily diverges.

But to prove the Statement we also have to Show that $ \sum_n 2^na_{2^n}$ diverges. It would be sufficient in this case to Show that $ (2^na_{2^n})$ is not Zero convergent.

How can I do that?