# Show $\sum_{n=0}^{\infty}b_k$ is absolutely convergent

$$\sum_{n=0}^{\infty}a_n$$ is abs. convergent

$$b_k =(a_0 +2a_1 +2^2a_2 +···+2^ka_k)2^{−(k+1)}$$

Show that $$\sum_{k=0}^{\infty}b_k$$ is abs. convergent and, $$\sum_{k=0}^{\infty}b_k = \sum_{k=0}^{\infty}a_n$$

I was said to use induction, but sincerely I don't find any useful way, is there any other approach?

• In the sums of $b_k$ it should be $k = 0$ instead of $n = 0$ I believe. – araomis Nov 8 '18 at 17:41
• @araomis yes sorry, I corrected – Dada Nov 8 '18 at 18:01

Let $$m \in N$$ be arbitrary. Consider the partial sum $$\Sigma_{k =0}^m b_k$$. By applying the definition of $$b_k$$ we get:
$$\Sigma_{k =0}^m b_k = \Sigma_{k =0}^m(a_0 +2a_1 +2^2a_2 +···+2^ka_k)2^{−(k+1)} = \Sigma_{k =0}^m((\Sigma_{i = 1}^{m - k + 1}2^{-i})a_k) \leq \Sigma_{k =0}^m a_k$$
Since $$\Sigma_{n = 0}^\infty a_n$$ is abs. convergent, it follows that $$\Sigma_{k = 0}^\infty b_k$$ must be abs. convergent.
Now you have to look at what happens for $$m \rightarrow \infty$$ in order to find out why the identity holds.