Absolute convergence of $\sum\limits_{k=0}^{\infty}2^{-(k+1)}\sum\limits_{j=0}^{k}2^ja_j$ 
Let be $\sum\limits_{n=0}^{\infty}a_n$ an absolutely convergent series and $b_k:=2^{-(k+1)}\sum\limits_{j=0}^{k}2^ja_j$. Show that $\sum\limits_{k=0}^{\infty}b_k$ converges absolutely.

 A: If $a_j$ are nonnegative real numbers, then we're free to change the order $\sum\limits_{k=0}^\infty\sum\limits_{j=0}^k=\sum\limits_{j=0}^\infty\sum\limits_{k=j}^\infty$: $$\sum_{k=0}^\infty 2^{-k-1}\sum_{j=0}^k 2^j a_j=\sum_{j=0}^\infty 2^j a_j\underbrace{\sum_{k=j}^\infty 2^{-k-1}}_{=2^{-j}}=\sum_{j=0}^\infty a_j.$$
For the general case, we use $|b_k|\leqslant 2^{-k-1}\sum\limits_{j=0}^k 2^j|a_j|$ and the above to get $\sum\limits_{k=0}^\infty|b_k|\leqslant\sum\limits_{j=0}^\infty|a_j|$.
A: I just want to share another answer which uses Cauchy's product rule (https://en.wikipedia.org/wiki/Cauchy_product#:~:text=In%20mathematics%2C%20more%20specifically%20in,French%20mathematician%20Augustin%20Louis%20Cauchy.).
We know that $\sum\limits_{j=0}^{\infty}a_j$ and $\sum\limits_{k=0}^{\infty}2^{-k-1}$ are both absolute convergent. Therefore we are allowed to apply Cauchy's product rule:
$1\cdot\sum\limits_{j=0}^{\infty}a_j=\sum\limits_{k=0}^{\infty}2^{-k-1}\sum\limits_{j=0}^{\infty}a_j=\sum\limits_{k=0}^{\infty}\sum\limits_{j=0}^{k}a_j2^{-(k-j)-1}=\sum\limits_{k=0}^{\infty}\sum\limits_{j=0}^{k}2^{j-(k+1)}a_j=\sum\limits_{k=0}^{\infty}b_k$.
Hence, $\sum\limits_{k=0}^{\infty}b_k$ converges absolutely and $\sum\limits_{k=0}^{\infty}b_k=\sum\limits_{j=0}^{\infty}a_j$.
