<s>[INST] Showing that $\sum\limits_{j=0}^\infty a_j=\sum\limits_{k=0}^\infty b_k$ Show when $\sum\limits_{j=0}^\infty a_j$ is absolute convergent, than $\sum\limits_{k=0}^\infty b_k$ with $b_k:=(a_0+2a_1+...+2^ka_k)/2^{k+1}$ is also absolute convergent,
and even $\sum\limits_{j=0}^\infty a_j=\sum\limits_{k=0}^\infty b_k$

My attempt:
$\sum\limits_{k=0}^\infty b_k=\frac{a_0}{2}+\left(\frac{a_0}{2^2}+\frac{2a_1}{2^2}\right)+\left(\frac{a_0}{2^3}+\frac{a_1}{2^3}+\frac{2^2a_2}{2^3}\right)+...=\sum\limits_{k=0}^\infty \left (\sum\limits_{i=0}^k \frac{1}{2^{k+1}}2^ia_i\right)$
$$\lim\limits_{k\rightarrow\infty}\left |\sqrt[k]{\sum\limits_{i=0}^{k} \left ( \frac{1}{2^{k+1}}2^ia_i\right)}\right|=0<1$$
Because  $\lim\limits_{k\rightarrow\infty}\left |\sum\limits_{i=0}^{k} \left ( \frac{1}{2^{k+1}}2^ia_i\right)\right|=\left |\sum\limits_{i=0}^{\infty} \left ( 0*2^ia_i\right)\right|=0$
$\sum\limits_{k=0}^\infty \left (\sum\limits_{i=0}^k \frac{1}{2^{k+1}}2^ia_i\right)=\sum\limits_{k=0}^\infty b_k$ is absolute convergent
Since $\sum\limits_{k=0}^\infty b_k$ is absolute convergent, we can rearrange its pieces:
$$\sum\limits_{k=0}^\infty b_k=\frac{a_0}{2}+\left(\frac{a_0}{2^2}+\frac{2a_1}{2^2}\right)+\left(\frac{a_0}{2^3}+\frac{2a_1}{2^3}+\frac{2^2a_2}{2^3}\right)+...=a_0\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...\right)+a_1\left(\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+...\right)+...$$
$$=a_0\left(\sum\limits_{k=1}^\infty \left(\frac{1}{2}\right)^k\right)+2a_1\left(\sum\limits_{k=2}^\infty \left(\frac{1}{2}\right)^k\right)+2^2a_2\left(\sum\limits_{k=3}^\infty \left(\frac{1}{2}\right)^k\right)+...$$
$=a_0\left(2-1\right)+2a_1\left(2-1-\frac{1}{2}\right)+2^2a_2\left(2-1-\frac{1}{2}-\frac{1}{4}\right)+...=a_0+\frac{1}{2}2a_1+\frac{1}{4}2^2a_2+...=\sum\limits_{k=0}^\infty a_k$
$\Box$

Hello could someone look over my work and give some feedback :)? Is my solution correct, and if not, what could I improve?
 [/INST] When you want to prove the absolute convergence of the series $\sum_k b_k$ you should prove that $\sum_k |b_k|$ is finite. When you only have non-negative terms, you can rearrange them in any order. Here
$$
\begin{aligned}
\sum\limits_{k=0}^\infty |b_k| &= \sum\limits_{k=0}^\infty \left |\sum\limits_{i=0}^k \frac{1}{2^{k+1}}2^ia_i\right|
\leq \sum\limits_{k=0}^\infty \sum\limits_{i=0}^k \frac{1}{2^{k+1}}2^i |a_i|
= \sum_{0 \leq i \leq k} \frac{1}{2^{k+1}}2^i |a_i| \\
&=\sum\limits_{i=0}^\infty 2^i |a_i| \underbrace{\left (\sum\limits_{k=i}^\infty \frac{1}{2^{k+1}}\right)}_{=\frac{1}{2^j}}
= \sum\limits_{i=0}^\infty |a_i| < \infty
\end{aligned}
$$
so the series $\sum_k b_k$ is absolutely convergent. Once you have proved this, you are guaranteed that you can rearrange the terms with $\sum\limits_{k=0}^\infty b_k$ so the same computations works, but you have an equality instead of an inequality in the second step so you get that $\sum_k b_k = \sum_k a_k$.
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