# prove that if $\sum 2^na_{2^n}$ converges then $\sum a_n$ converges

Suppose $\sum a_n$ is a series with decreasing, nonnegative terms. Define $$s_n = \sum_{k=1}^n a_k = a_1+\cdots +a_n$$ and $$t_n = \sum _{k=0}^n 2^ka_{2^k} = a_1+2a_2+4a_4+\cdots +2^na_{2^n}.$$

Also, $s_n$ is defined for all $n\geq 1$ whereas $t_n$ is defined for all $n\geq 0$. I have already proved that $$s_{2^{n+1}-1}\leq t_n$$ for all $n\geq 0$.

Proof: Since $a_n$ is decreasing then $2a_2>a_2$ and $4a_4>a_4+a_5+a_6+a_7$ so then $2^na_{2^n} > a_{2^2}+...+a_{2^{n+1}-1}$ so therefore $s_{2^{n+1}-1}$ for all $n\geq0$.

However now I need to use that proof I have found above and try and prove that if $\sum 2^na_{2^n}$ converges then $\sum a_n$ converges. Looking for some help with this second proof.

• @Rafael Wagner thanks for the edit, do you know how to work through this proof? – hburt Jan 20 '17 at 17:03
• $t_n \leqslant -a_1 + 2\bigl( a_1 + a_2 + (a_3 + a_4) + (a_5 + a_6 + a_7 + a_8) + \dotsc\bigr)$ – Daniel Fischer Jan 20 '17 at 17:06
• does that prove $\sum 2^na_{2^n}$ converges then $\sum a_n$ converges – hburt Jan 20 '17 at 17:08
• No, that's the other part of the "if and only if". For the direction you want, the inequality you already have in your question is enough. Together with something about monotonic sequences. – Daniel Fischer Jan 20 '17 at 17:10
• @hburt the answer given below is a correct answer. Can you understand it ? – Rafael Wagner Jan 20 '17 at 19:27

Notice that $s_w=\sum\limits_{n=1}^w a_m$ is a non-decreasing sequence. Take any $w$ and let $2^m> w$, then:
$s_w=\sum\limits_{n=1}^w a_n\leq \sum\limits_{n=1}^{2^m-1}a_n \leq \sum\limits_{n=1}^{m-1} a_{2^n}+a_2+\dots+ a_{2^{n+1}-1}\leq \sum\limits_{n=1}^{m-1} 2^na_{2^n}\leq \sum\limits_{n=1}^\infty 2^n a_{2^n}$. We have proved $s_n$ is non-decreasing and bounded, and hence converges.
Analogously, if $\sum\limits_{n=1}^\infty a_n$ converges we have:
$\frac{\sum\limits_{n=1}^w 2^n a_{2^n}}{2}\leq \sum \limits_{n=1}^w a_{2^{n-1}+1}+a_{2^{n-1}+2}+\dots a_{2^n}\leq \sum \limits_{n=1}^\infty a_n$.