# Dyadic series proof?

Prove that if the the terms of a sequence decrease monotonically, $a_1 \geq a_2 \geq a_3 \geq \dots \geq a_n$, and converge to 0 then the series $\Sigma a_k$ converges iff the associated dyadic series $a_1 + 2a_2 + 4a_4 + \dots + = \Sigma 2^k a_{2^k}$ converges.

Given: $a_1 \geq a_2 \geq a_3 \geq \dots \geq a_n$, and converges to 0. \ \textbf{Part 1:} \ {Assume} $a_1 + 2a_2 + 4a_4 + \dots + = \Sigma 2^k a_{2^k}$ is convergent. Then, we have: \ $a_1 \leq a_1$ \ $a_2 + a_3 \leq a_2 + a_2 = 2a_2$ \ $a_4 + a_5 + a_6 + a_7 \leq a_4 + a_4 + a_4 + a_4 = 4a_4$ \ $a_8 + \dots + a_{15} \leq 8a_8$ \ \noindent Adding the inequalities above, we have: \ $\Sigma_{n=1}^\infty a_n \leq \Sigma_{k=0}^\infty 2^k a_{2^k}$ \ Thus, since the RHS is convergent and all the terms are positive, by the Comparison Test, the LHS must also be convergent. Hence, $\Sigma_{n=1}^\infty a_n$ is convergent. \ \textbf{Part 2:} \ {Assume} $\Sigma_{n=1}^\infty a_n$ converges. Then, $a_1 + 2 \Sigma_{n=1}^\infty a_n$ also converges since we are merely scaling by a constant and adding a constant term. \ Recall that $a_1 \geq a_2 \geq a_3 \geq \dots \geq a_n$ Then, \ $a_1 \leq a_1$ \ $2a_2 \leq 2a_1$ \ $4a_4 \leq 2a_2 + 2a_3$\ $8a_4 \leq 2a_4 + 2a_5 + 2a_6 + 2a_7$\ Adding the inequalities above, we have: \ $\Sigma_{k=0}^\infty 2^k a_{2^k} \leq a_1 + 2 \Sigma_{n=1}^\infty a_n$\ Since the RHS converges and all elements are positive, the LHS must converge by the Comparison Test. Hence, $\Sigma_{k=0}^\infty 2^k a_{2^k}$ converges.