Proof of the Cauchy condensation test

Proof

We compare the two sequences of partial sums by grouping terms in blocks of length a power of $2$.

1. For $n \ge 1$ let $s_n = \sum_{m=1}^{n} a_m$, and for $K \ge 0$ let $t_K = \sum_{k=0}^{K} 2^k a_{2^k}$. These are the partial sums of the two series. Both have nonnegative terms, so both sequences are increasing.
2. By the criterion for nonnegative series, each series converges if and only if its partial sums are bounded above. A shift of the starting index does not change this.
3. Let $k \ge 0$. The block of indices $m$ with $2^k \le m \le 2^{k+1} - 1$ has $2^k$ elements. Since $(a_n)$ is decreasing, as a monotone sequence, each term in it satisfies $a_m \le a_{2^k}$.
4. So the block sum is at most $2^k a_{2^k}$. Adding over $k = 0, \dots, K$ gives $s_{2^{K+1} - 1} \le t_K$.
5. Let $k \ge 1$. The block of indices $m$ with $2^{k-1} < m \le 2^k$ has $2^{k-1}$ elements, and each term in it satisfies $a_m \ge a_{2^k}$.
6. So that block sum is at least $2^{k-1} a_{2^k}$. Adding over $k = 1, \dots, K$ and adding $a_1$ gives $s_{2^K} \ge a_1 + \sum_{k=1}^{K} 2^{k-1} a_{2^k}$.
7. The right side of step 6 is at least $\tfrac12 a_1 + \tfrac12 \sum_{k=1}^{K} 2^k a_{2^k} = \tfrac12 t_K$, since $a_1 \ge 0$. So $t_K \le 2 s_{2^K}$.
8. Assume that the condensed series converges, with sum $T$. Then $t_K \le T$ for all $K$, by step 2. Let $n \ge 1$. Since $n \le 2^{n+1} - 1$ and $(s_n)$ is increasing, step 4 gives $s_n \le s_{2^{n+1}-1} \le t_n \le T$.
9. So $(s_n)$ is bounded above, and $\sum a_n$ converges by step 2.
10. Assume that $\sum a_n$ converges, with sum $S$. Then $s_n \le S$ for all $n$. By step 7, $t_K \le 2 s_{2^K} \le 2S$ for all $K$.
11. So $(t_K)$ is bounded above, and the condensed series converges by step 2.