$$\sum_{n=0}^{\infty}a_n$$ 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?

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


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