14
$\begingroup$

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?

  • 0
    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](http://imperator.usc.edu/~bruck/classes/fall2009/math425a/Cauchy_condensation.pdf).2012-12-17

2 Answers 2

15

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. $\Box$

  • 1
    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 $\forall k\in \Bbb N$.2016-06-02
3

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}. $