Is the following argument correct?
Proposition. Given that $(b_n)$ is a decreasing sequence such that $b_n\ge 0,\forall n\ge 0$. Prove that the series $\sum_{n=1}^{\infty}b_n$ diverges whenever $\sum_{n=1}^{\infty}2^nb_{2^n}$ diverges.
Proof. Assume that $\sum_{n=1}^{\infty}2^nb_{2^n}$ diverges, then given any $M>0$, there exists a $k\in\mathbf{N}$ such that $$\sum_{n=1}^{k}2^nb_{2^n}\ge M$$ from hypothesis we know that $(b_n)$ is decreasing consequently we have $$\sum_{n=1}^{k}2^nb_n\ge\sum_{n=1}^{k}2^nb_{2^n}\ge M$$ In addition it is not difficult to see that $1+2+2^2+2^4+\cdots+2^{k} = 2^{k+1}-1$ consequently $$(2^{k+1}-1)\sum_{n=1}^{k}b_n \ge\sum_{n=1}^{k}2^nb_n\ge\sum_{n=1}^{k}2^nb_{2^n}$$ implying $$\sum_{n=1}^{k}b_n\ge \frac{M}{2^{k+1}-1}$$ given this it follows that the series $\sum_{n=1}^{\infty}b_n$ can be made arbitrarily large and thus diverges.
$\blacksquare$
No, there are several errors. The equation $(2^{k+1}-1)\sum_{n=1}^{k}b_n = \sum_{n=1}^{k}2^nb_n$ is simply not right. More important, the overall outline is not clear - this looks like a proof that $\sum 2^n b_n$ diverges.
Edit: It turns out that that equation was a typo for $(2^{k+1}-1)\sum_{n=1}^{k}b_n \ge \sum_{n=1}^{k}2^nb_n$. Fine, but it doesn't matter, because the argument still does not prove what needs to be proved! What we have is a proof that $\lim_k (2^{k+1}-1)\sum_{n=1}^{k}b_n=\infty$. That does not imply $\lim_k\sum_1^k b_n=\infty$, (which is what we need to show), because $\lim_k 2^{k+1}=\infty$.
Or to put it another way: The final "given this it follows" does not follow, because $M/(2^{k+1}-1)\to0$ as $k\to\infty$.
Hint for a correct proof: $$b_3+b_4\ge b_4+b_4=\frac12 4b_4,$$ $$b_5+b_6+b_7+b_8\ge 4b_8=\frac12 8b_8,$$ etc.