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?
 A: If $(a_n)_{n\geqslant1}$ is nonnegative and nonincreasing, then
$$
a_1+\sum_{n=0}^{+\infty}2^na_{2^n}\leqslant2\sum_{n=1}^{+\infty}a_n\leqslant2\sum_{n=0}^{+\infty}2^na_{2^n}.
$$
A: 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. $\Box$
