Convergence of monotone decreasing series $\sum_{n=1}^\infty a_n < \infty \iff \sum_{n=1}^\infty 2^na_{2^n} < \infty$ Suppose $\{a_n\}$ is a monotone decreasing sequence of positive terms.  Prove that $$\sum_{n=1}^\infty a_n \text{ converges } \iff \sum_{n=1}^\infty 2^na_{2^n} \text{ converges}$$
thought about the integral test but there's no function to integrate, can't assume $a_n\to 0$ implies convergence, and having a tough time using comparison.  Would ratio test do it? 
 A: Observe that
$$\begin{aligned}\sum_{n=1}^\infty a_n&=a_1+a_2+a_3+\dots\\
&=a_1+(a_2+a_3)+(a_4+a_5+a_6+a_7)+\dots\\
&\leq a_1+(a_2+a_2)+(a_4+a_4+a_4+a_4)+\dots\\
&=a_1+2a_2+4a_4+\dots\end{aligned}$$
and
$$\begin{aligned}\sum_{n=1}^\infty 2^na_{2^n}&=a_1+2a_2+4a_4+\dots\\
&=(a_1+a_2)+(a_2+a_4+a_4+a_4)+\dots\\
&\leq (a_1+a_1)+(a_2+a_2+a_3+a_3)+\dots\\
&=2\sum_{n=1}^\infty a_n\end{aligned}$$
A: Proposition 1 (Cauchy criterion). Let $(a_n)_{n=1}^\infty$ be a decreasing sequence of non-negative real numbers (so $a_n\ge0$ and $a_{n+1}\le a_n$ for all $n\ge 1$). Then the series $\sum_{n=1}^\infty a_n$ is convergent if and only is the series $$\sum_{k=0}^\infty2^ka_{2^k}=a_1+2a_2+4a_4+8a_8+\dotsb$$ is convergent.
Lemma 2. Let $\sum_{n=m}^\infty a_n$ be a series of non-negative real numbers. Then this series is convergent if and only if there is a real number $M$ such that $$\sum_{n=m}^Na_n\le M\;\;\text{for all integers}\;\;N\ge m.$$
Proof of Proposition 1. Let $S_N:=\sum_{n=1}^Na_n$ be the partial sums of $\sum_{n=1}^\infty a_n$, and let $T_K:=\sum_{k=1}^K2^ka_{2^k}$ be the partial sums of $\sum_{k=0}^\infty 2^ka_{2^k}$. In light of Lemma 2, our tak is to show that the sequence $(S_N)_{N=1}^\infty$ is bounded if and only if the sequence $(T_K)_{K=0}^\infty$ is bounded.
Lemma 3. For any natural $K$, we have $S_{2^{K+1}-1}\le T_K\le 2S_{2^K}$
From Lemma 3 we see that if $(S_N)_{N=1}^\infty$ is bounded, then $(S_{2^K})_{K=0}^\infty$ is bounded, and hence $(T_K)_{K=0}^\infty$ is bounded.
Conversely, if $(T_K)_{K=0}^\infty$ is bounded, then the Lemma 3 implies that $S_{2^{K+1}-1}$ is bounded, i.e., there is an $M$ such that $S_{2^{K+1}-1}\le M$ for all natural numbers $K$. But one can esily show (using induction) that $S_{2^{K+1}-1}\ge K+1$, and hence that $S_{K+1}\le M$ for all natural numbers $K$, hence $(S_N)_{N=1}^\infty$ is bounded.
