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I'm trying to show that then $\Sigma a_n$ converges iff $\Sigma 2^n a_{2^n}$ converges if $a_n$ is any monotone sequence.

I am assuming I have to prove the cases:

  • $a_n$ is positive and increasing
  • $a_n$ is negative and increasing
  • $a_n$ is positive and decreasing
  • $a_n$ is negative and decreasing

Here is what I have worked out from some of the comments below:

Assume $a_n$ is an increasing sequence. If $a_n$ is positive, then $\lim a_n \neq 0$ and we have divergence.

I don't really understand why this is. Is it because $a_n$ is unbounded and so it diverges to $\infty$? I don't know a rigorous way to prove this. I'm assuming I have to make use of the theorem that monotonic $s_n$ converges iff it is bounded.

If $a_n$ is negative, then we have that $\lim a_n = 0$

Again, I don't know how to rigorously prove this. How do I find a bound for $a_n$?

Let $s_n = a_1 + a_2 + ... + a_n$ and $t_n = a_1 + 2 a_2 + ... + 2^k a_{2^k}$. For $n<2^k$, we have $s_n \geq a_1 + (a_2 + a_3) + ... _ (a_{2^k}+...+a_{2^{k+1}-1} \geq a_1 + a_2 + ... + 2^{k}a_{2k} = t_k $

And so $s_n \geq t_k$.

Similar proof for $2s_{n} \leq t_k$

Someone below has already proved nonnegative decreasing case. I'm assuming that:

If $a_n$ was decreasing and negative, then $\lim a_n \neq 0$ and we have divergence.

Again, I'm not sure how to prove this.

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Consider the series $\sum a_n $ where $(a_n)$ is decreasing and positive. Also, consider the series $\sum 2^n a_{2^n}$. Consider their partial sums $(s_n), (t_n)$

Note if $n < 2^n$, then $s_n = a_1 + a_2 + ... + a_n \leq a_1 + (a_2+a_3) + (a_4 + a_5 + a_6 + a_7) + ... + (a_{2^n} + ... + a_{2^{n+1}-1}) \leq a_1 + 2a_2 + ... + 2^n a_{2n} = t_n \implies \boxed{ s_n \leq t_n }$

If $2^n < n$, then $s_n \geq a_1 + a_2 + (a_3 + a_4) + ... + (a_{2^{n-1}+1} + ... + a_{2^n} ) \geq \frac{1}{2} a_1 + a_2 + 2 a_4 + ... + 2^{n-1}a_n = \frac{1}{2} t_n \implies \boxed{ 2s_n \geq t_n }$

Since series of nonnegative terms converges iff its sequence of partial sums are bounded, then we are done.

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  • $\begingroup$ Do we have the consider when $a_n$ is increasing and positive? I think someone below explained why negative terms will result in a divergence. Sorry if this seems like a dumb question. $\endgroup$ – Nikitau Nov 9 '16 at 1:59
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Plenty of information at the Wikipedia page on this very theorem:

http://en.wikipedia.org/wiki/Cauchy_condensation_test

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  • $\begingroup$ I appreciate the link, but it states that the test holds true for nonnegative, decreasing sequences. I want to somehow show that it's true for any monotone sequence, but I don't see how this could be true given the conditions for the proof of Cauchy Condensation Test. $\endgroup$ – Nikitau Nov 10 '16 at 20:34
  • $\begingroup$ If the sequence is monotone then $a_n \to 0$ and $a_{2^n} \to 0$ are equivalent. So the only cases that potentially converge are negatives of each other (positive decreasing, and negative increasing, both converging to 0). $\endgroup$ – zyx Nov 11 '16 at 4:22
  • $\begingroup$ Could you explain why the positive increasing would not result in convergence? It might be a silly question, but I'm not sure how to prove it. $\endgroup$ – Nikitau Nov 11 '16 at 5:35
  • $\begingroup$ The terms of the sum have to go to 0 in order for the sum to converge. $\endgroup$ – zyx Nov 11 '16 at 5:41
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This is called the Cauchy Condensation Test. Note the convergence or divergence of $\sum_na_n$ is unaffected by changing $a_n$ for any finite set of $n.$ So if $a_n<0$ for only finitely many n, we may substitute $0$ for $a_n$ whenever $a_n<0$ without affecting convergence or divergence. (Similarly if $a_n>0$ for finitely many $n.$)

Also it is sufficient that $(a_n)_n$ is monotonic for all but finitely many n. ... If $(a_n)_n$ is decreasing (for all but finitely many $n$) but has negative values for infinitely many $n$ then for some $r>0$ we have: $a_n<-r$ for all but finitely many $n$ so $\sum_na_n$ diverges, as we do not even have $a_n\to 0.$ ... I see that a proof of the Test has appeared as an answer just now.

Examples of application:

(1). Let $a_n=n^{-K}$ for each $n.$ Then $\sum_na_n$ converges iff $K>1.$

(2).$\sum_{n>1}\frac {1}{n(\log n)^K}$ converges iff $K>1.$

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  • $\begingroup$ Thanks for the explanation! This might be a dumb question, but do we have the consider the case where the sequence is positive and increasing? $\endgroup$ – Nikitau Nov 9 '16 at 2:00
  • $\begingroup$ If $(a_n)_n$ is positive and increasing then $\sum_{j=1}^na_j\geq na_1, $ which goes to $\infty$ as $n\to \infty,$ because $a_1>0.$ But the Cauchy Condensation Test works for all monotone sequences. $\endgroup$ – DanielWainfleet Nov 9 '16 at 6:14
  • $\begingroup$ Thanks, but I still think I'm a little confused. Indeed, I am trying to prove that the Cauchy Condensation Test works for any monotone sequences. So in my proof, do I need to include the cases where we have a positive increasing, negative increasing, positive decreasing and negative decreasing sequence? What if we have a sequence with both positive and negative terms? $\endgroup$ – Nikitau Nov 9 '16 at 23:31

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