Induction on a series with absolute value

Question

$$\sum_{k=0}^{n}|b_k|≤\sum_{k=0}^{n}|a_k|.(1-\frac{1}{2^{n+1-k}})$$ $b_k=(a_0+2a_1+2^2a_2+…+2^ka_k)2^{-(k+1)}$

I have to prove this inequality with absolute value with induction, but I don't know how to treat this with ≤

Answer 1

Using the triangle inequality, we have that \begin{align*} \sum\limits_{k=0}^n |b_k| &= \sum\limits_{k=0}^n \frac{1}{2^{k+1}}|a_0+2a_1+ \cdots + 2^{k-1}a_{k-1}+ 2^ka_k|\\ &\leq \sum\limits_{k=0}^n \frac{1}{2^{k+1}}\left(|a_0|+|2a_1|+\cdots + |2^{k-1}a_{k-1}|+ |2^ka_k|\right)\\ &= \sum\limits_{k=0}^n \frac{|a_0|}{2^{k+1}} + \sum\limits_{k=0}^{n-1} \frac{|a_1|}{2^{k+1}} + \cdots + \sum\limits_{k=0}^1 \frac{|a_{n-1}|}{2^{k+1}} + \frac{|a_n|}{2}\\ &= |a_0|\left(1-\frac{1}{2^{n+1}} \right) + |a_1|\left(1-\frac{1}{2^{n}} \right) + \cdots + |a_{n-1}|\left(1-\frac{1}{2^{2}} \right)+|a_{n}|\left(1-\frac{1}{2} \right)\\ &= \sum\limits_{k=0}^n |a_k|\left(1-\frac{1}{2^{n+1-k}}\right) \end{align*}