Proving Cauchy condensation test

Solution 1:

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. $\Box$

Solution 2:

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