My task is this:

(i) Let $\{a_n\}$ be a positive decreasing sequence.

Show that:$$2^ma_{2^m}\geq a_{2^m} + a_{2^m+1}+a_{2^m+2}+\ldots+a_{2^{m+1}+1}\geq2^ma_{2^{m+1}}.$$

(ii) Use this to show that:

$\sum_{n=0}^\infty a_n$ converges iff $\sum_{m=0}^\infty 2^ma_{2^m}$ .

This is called Cauchy condensation test.

My work:

Since $\{a_n\}$ is positive and decreasing we must have that $a_{2^m}\geq a_{2^{m+1}} \implies 2^ma_{2^m}\geq 2^ma_{2^{m+1}}$. Now for the expression in the middle it should be easy to see that: $$\begin{align}a_1 &\geq a_1. \\2a_2&\geq a_2 + a_3.\\ 4a_4&\geq a_4 + a_5 + a_6 + a_7.\\ & \ \vdots\\ 2^ma_{2^m} &\geq a_{2^m} + a_{2^m+1}+a_{2^m+2}+\ldots+a_{2^{m+1}-1}.\end{align} $$

If this wasn't true, it woudn't be a decreasing sequence. Now this is where the troublesome part comes into play for me atleast since if $\sum_{n=0}^\infty a_n$ is to converge iff $\sum_{m=0}^\infty 2^ma_{2^m}$ converges, shouldn't that imply that $\sum_{n=0}^\infty a_n \leq \sum_{m=0}^\infty 2^ma_{2^m}$? If yes then this is what we get when writing this one out:$$a_0 + a_1 + a_2 + \ldots+a_n +\ldots\leq a_1 +2a_2 + 4a_4+\ldots+2^ma_{2^m}+\ldots$$

How can we be sure that this is true? Consider the case where the terms are equal from the inequalities above, then we have a term that's greater than or equal to anyone of the other terms on the right side, namely $a_0$.

Thanks in advance.

  • $\begingroup$ The first term in $\sum_{m=0}^\infty2^ma_{2^m}$ is $a_1$, not $a_0$. $\endgroup$ – Barry Cipra May 13 '16 at 20:24
  • $\begingroup$ @Barry Cipra Yes thanks alot for pointing this out. I've added the changes now. $\endgroup$ – Thomas May 14 '16 at 7:01

I think your problem is with the limits of summation in the Cauchy condensation test. It seems as if wikipedia lists the test as $\sum_{n=1}^\infty a_n \leq \sum_{m=0}^\infty 2^ma_{2^m}$ and you wrote $\sum_{n=0}^\infty a_n \leq \sum_{m=0}^\infty 2^ma_{2^m}$ Furthermore, it appears as you prove the equation listed on wikipedia.

In the case you tell me to consider, where $$ \begin{align} a_1 &= a_1 \\ 2a_2&= a_2 + a_3\\ 4a_4&= a_4 + a_5 + a_6 + a_7\\ &\ \ \vdots\\ 2^ma_{2^m} &= a_{2^m} + a_{2^m+1}+a_{2^m+2}+\ldots+a_{2^{m+1}-1}. \end{align} $$

this simplifies to

$$ \begin{align} 0 &= 0 \\ a_2&= a_3\\ 3a_4&= a_5 + a_6 + a_7\\ &\ \ \vdots\\ (2^m-1)a_{2^m} &= a_{2^m+1}+a_{2^m+2}+\ldots+a_{2^{m+1}-1}. \end{align} $$

summing these terms as you did gives

$$a_3 + a_5 + a_6 + a_7 + \ldots + \sum_{n=2^m+1}^{2^{m+1}}a_n = a_2 + 3a_4 + \ldots + (2^m-1)a_{2^m}$$

I see no contradiction here, as if you group the terms the way there were grouped originally in the inequalities, the final equality is a direct result.

Also, generalizing to the case of inequality again, the final formula (now with an inequality in it) is fairly clearly a result of it being a positive decreasing sequence.

  • $\begingroup$ That is probably right, it could be a typo. However the book that I got this exercise from certainly states the cases for $n=m=0$. I'll ask the professor who wrote the book or someone working under him if I get the chance next week. $\endgroup$ – Thomas May 14 '16 at 12:11

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