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.
Absolute convergence of $\sum\limits_{k=0}^{\infty}2^{-(k+1)}\sum\limits_{j=0}^{k}2^ja_j$
convergence-divergencereal-analysissequences-and-seriessolution-verification
Related Solutions
If the series $\sum_{n=1}^\infty |a_n| \sum_{k=n}^\infty |b_k| $ converges, then you can do whatever you want with $a_n b_k$; any rearrangement will converge to the same sum. More generally: if $I$ is an index set and $\sum_{i\in I} |c_i|<\infty$, then any rearrangement of $c_i$ converges to the same sum. This is because for every $\epsilon>0$ there is a finite set $S\subset I$ such that $\sum_{i\in I\setminus S}|c_i|<\epsilon $; any method of summation will use up $S$ at some point, and after that the sum is guaranteed to be within $\epsilon$ of $\sum_{i\in S} c_i$.
But if you only know that $$\sum |b_k|\tag{1}$$ and $$\sum_{n=1}^\infty \left|a_n \sum_{k=n}^\infty b_k \right|\tag{2} $$ converge, then rearrangements can go wrong. For example, take the series $\sum b_k $ to be $\frac{1}{2} -\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+\frac{1}{8}-\frac{1}{8}+\dots$ then every other sum $\sum_{k\ge n} b_k$ is zero, and the corresponding $a_n$ could be arbitrarily large without disturbing the convergence of (2). In this situation you could even have infinitely many terms $a_nb_k$ that are greater than $1$ in absolute value; clearly this series can't be rearranged at will.
First assume that $a_n \ge 0$ and define $\sum_{n \in I} a_n = \sup_{J \subset I, J \text{ finite}} \sum_{n \in J} a_n$. Note that it follows that if $I \subset I'$ then $\sum_{n \in I} a_n \le \sum_{n \in I'} a_n$.
From https://math.stackexchange.com/a/3680889/27978 we see that if $K = K_1 \cup \cdots \cup K_m$, a disjoint union, then $\sum_{n \in K} a_n = \sum_{n \in K_1} a_n + \cdots + \sum_{n \in K_m} a_n$.
Since $I'=I_1 \cup \cdots \cup I_m \subset I$ we see that $\sum_{n \in I} a_n \ge \sum_{n \in I'} a_n = \sum_{k=1}^m \sum_{n \in I_k} a_n$. It follows that $\sum_{n \in I} a_n \ge \sum_{k=1}^\infty \sum_{n \in I_k} a_n$. This is the 'easy' direction.
Let $\epsilon>0$, then there is some finite $J \subset I$ such that $\sum_{n\in J} a_n > \sum_{n \in I} a_n -\epsilon$. Since $J$ is finite and the $I_k$ are pairwise disjoint we have $J \subset I'=I_1 \cup \cdots \cup I_m$ for some $m$ and so $\sum_{k=1}^\infty \sum_{n \in I_k} a_n \ge \sum_{k=1}^m\sum_{n \in I_k} a_n \ge \sum_{k=1}^m\sum_{n \in J \cap I_k} a_n = \sum_{n\in J} a_n > \sum_{n \in I} a_n -\epsilon$.
(It is not relevant here, but a small proof tweak shows that the result holds true even if the $a_n$ do not have a finite sum.)
Now suppose we have $a_n \in \mathbb{R}$ and $\sum_{n \in I} |a_n| = \sum_{n=1}^\infty |a_n|$ is finite. We need to define what we mean by $\sum_{n \in I} a_n$. Note that $(a_n)_+=\max(0,a_n) \ge 0$ and $(a_n)_-=\max(0,-a_n) \ge 0$. Since $0 \le (a_n)_+ \le |a_n|$ and $0 \le (a_n)_- \le |a_n|$ we see that $\sum_{n \in I} (a_n)_+ = \sum_{k=1}^\infty \sum_{n \in I_k} (a_n)_+$ and similarly for $(a_n)_-$.
This suggests the definition (cf. Lebesgue integral) $\sum_{n \in I} a_n = \sum_{n \in I} (a_n)_+ - \sum_{n \in I} (a_n)_-$.
With this definition, all that remains to be proved is that $\sum_{k=1}^\infty \sum_{n \in I_k} a_n = \sum_{k=1}^\infty \sum_{n \in I_k} (a_n)_+ - \sum_{k=1}^\infty \sum_{n \in I_k} (a_n)_-$ and this follows from summability and the fact that for each $k$ we have $\sum_{n \in I_k} a_n = \sum_{n \in I_k} (a_n)_+ - \sum_{n \in I_k} (a_n)_-$.
Note: To elaborate the last sentence, recall that I defined $\sum_{n \in I_k} a_n$ to be $\sum_{n \in I_k} (a_n)_+ - \sum_{n \in I_k} (a_n)_-$, so all that is happening here is the definition is applied to $I_k$ rather than $I$. Then to finish, note that if $d_k,b_k,c_k$ are summable and satisfy $d_k=b_k-c_k$ then $\sum_{k=1}^\infty d_k= \sum_{k=1}^\infty b_k- \sum_{n=1}^\infty c_k$, where $d_k = \sum_{n \in I_k} a_n$, $b_k = \sum_{n \in I_k} (a_n)_+$ and $c_k = \sum_{n \in I_k} (a_n)_-$.
Best Answer
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 $\sum\limits_{k=0}^\infty|b_k|\leqslant\sum\limits_{j=0}^\infty|a_j|$.