1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ @Rafael Wagner thanks for the edit, do you know how to work through this proof? $\endgroup$ – hburt Jan 20 '17 at 17:03
  • $\begingroup$ $t_n \leqslant -a_1 + 2\bigl( a_1 + a_2 + (a_3 + a_4) + (a_5 + a_6 + a_7 + a_8) + \dotsc\bigr)$ $\endgroup$ – Daniel Fischer Jan 20 '17 at 17:06
  • $\begingroup$ does that prove $\sum 2^na_{2^n}$ converges then $\sum a_n $ converges $\endgroup$ – hburt Jan 20 '17 at 17:08
  • $\begingroup$ 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. $\endgroup$ – Daniel Fischer Jan 20 '17 at 17:10
  • 1
    $\begingroup$ @hburt the answer given below is a correct answer. Can you understand it ? $\endgroup$ – Rafael Wagner Jan 20 '17 at 19:27
2
$\begingroup$

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:

$\frac{\sum\limits_{n=1}^w 2^n a_{2^n}}{2}\leq \sum \limits_{n=1}^w a_{2^{n-1}+1}+a_{2^{n-1}+2}+\dots a_{2^n}\leq \sum \limits_{n=1}^\infty a_n$.

So if either series converges, so does the other one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.