Proving Cauchy condensation test

I have to prove the condensation test of Cauchy by tomorrow and I am really unconfident about what I did:

$$\sum_{n=1}^\infty a_n\text{ converges } \iff \sum_{n=1}^\infty 2^n a_{2^n}\text{ converges}$$

I did the following:

Let $(b_n)$ be a sequence as follow: $b_{2^k+m}:=a_{2^k}$ with $k\in\mathbb N_0$ and $0\leq m<2^k$.

It's $a_{n+1}\leq a_n$ and so $0\leq a_{n+p}\leq a_n$ for all $n,p\in\mathbb N$.

So $\sum\limits_{n=1}^\infty b_n$ converges by the majorizing series $\sum\limits_{n=1}^\infty a_n$. And it's $\sum\limits_{n=0}^\infty b_n=\sum\limits_{n=0}^\infty\sum\limits_{m=0}^{2^n-1}a_{2^{n+1}}=\sum\limits_{n=1}^\infty 2^{n-1}a_{2^n}$ so $\Rightarrow$ is done.

For $\Leftarrow$ consider $c_{2^k+m}:=a_{2^k}$ with $k\in\mathbb N_0$ and $0\leq m<2^k$.

It's $|a_n|\leq c_n$ and $\sum\limits_{n=0}^\infty c_n=\sum\limits_{n=0}^\infty\sum\limits_{m=0}^{2^n-1}a_{2^{n}}=\sum\limits_{n=1}^\infty 2^{n}a_{2^n}$ and so $\sum\limits_{n=1}^\infty a_n$ converges by the majorizing series $\sum\limits_{n=0}^\infty c_n$.

Is this in form and content correct?

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Looks good to me. – process91 Dec 16 '12 at 19:10
@MichaelBoratko no doubts? everything is right? the most important question to me is if you really can handle the sums as above... – user46021 Dec 16 '12 at 19:31
You can rearrange/group terms of an absolutely convergent series, and here you prove (absolute) convergence before doing so, therefore it looks alright. – process91 Dec 16 '12 at 19:34
Actually, your $\Leftarrow$ argument needs to be slightly modified - write the string of series equalities in the other direction, since we know that the series $\sum 2^n a_{2^n}$ converges then you can ungroup the terms in the way that you want to. – process91 Dec 16 '12 at 19:38
Others have answered, but in case this could also be of use, here's an old handout of mine (written in 1998) at one of Ronald Bruck's webpages: The Cauchy Condensation Test. – Dave L. Renfro Dec 17 '12 at 21:30

Let $a_n$ be nonincreasing and nonnegative (this just follows from $a_{n+1}\leq a_n$) Now we will use the comparison test:
Let $\sum_{n=1}^\infty a_{n}$ be convergent. We get \begin{align*} a_1+\frac12\sum_{n=1}^K2^na_{2^n}&=a_1+a_2+2a_4+4a_8+\dots+2^{K-1}a_{2^K}\\ &\leq a_1+ a_2+a_3+a_4+a_5+a_6+a_7+a_8+\dots+a_{2^{K-1}+1}+\dots+a_{2^K-1}+a_{2^K}\\ &\leq \sum_{n=1}^\infty a_n \end{align*} So its partial sums are bounded and $\sum_{n=1}^\infty 2^na_{2^n}$ is convergent.
Now let $\sum_{n=1}^{\infty}2^na_{2^n}$ be convergent. We get \begin{align*} \sum_{n=1}^Na_n&=a_1+a_2+a_3+a_4+\dots+a_N\\ &\leq a_1+\left(a_2+a_3\right)+\left(a_4+a_5+a_6+a_7\right)+\dots+\left(a_{2^N}+\dots+a_{2^{N+1}-1}\right)\\ &\leq a_1+2a_2+4a_4+\dots+2^Na_{2^N}\\ &\leq a_1+\sum_{n=1}^\infty2^na_{2^n}\end{align*} And so the other sum converges. $\Box$
As a minor nitpick, I agree with the notation $\sum\limits_{n=1}^\infty <\infty$ for something converging, but $\sum\limits_{n=1}^K<\infty$ for all partial sums are bounded is a bit of a stretch. Surely leaving it at the line before makes more sense, since $\sum\limits_{n=1}^k\frac{1}{n}<\infty$ is true $\forall k\in \Bbb N$. – snulty Jun 2 at 14:31
If $(a_n)_{n\geqslant1}$ is nonnegative and nonincreasing, then $$a_1+\sum_{n=0}^{+\infty}2^na_{2^n}\leqslant2\sum_{n=1}^{+\infty}a_n\leqslant2\sum_{n=0}^{+\infty}2^na_{2^n}.$$