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  "body": "\n## Proof\n\nWe compare the two sequences of partial sums by grouping terms in blocks of length a power of $2$.\n\n1. 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|0NB643K60Q]] of the two series. Both have nonnegative terms, so both sequences are increasing.\n2. By the [[criterion for nonnegative series|0MX1HZBBXJ]], each series converges if and only if its partial sums are bounded above. A shift of the starting index does not change this.\n3. 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|0KN22AAW5R]], each term in it satisfies $a_m \\le a_{2^k}$.\n4. So the block sum is at most $2^k a_{2^k}$. Adding over $k = 0, \\dots, K$ gives $s_{2^{K+1} - 1} \\le t_K$.\n5. Let $k \\ge 1$. The block of indices $m$ with $2^{k-1} \u003c m \\le 2^k$ has $2^{k-1}$ elements, and each term in it satisfies $a_m \\ge a_{2^k}$.\n6. So that block sum is at least $2^{k-1} a_{2^k}$. Adding over $k = 1, \\dots, K$ and adding $a_1$ gives $s_{2^K} \\ge a_1 + \\sum_{k=1}^{K} 2^{k-1} a_{2^k}$.\n7. 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 $t_K \\le 2 s_{2^K}$.\n8. 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 $s_n \\le s_{2^{n+1}-1} \\le t_n \\le T$.\n9. So $(s_n)$ is bounded above, and $\\sum a_n$ converges by step 2.\n10. 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 $K$.\n11. So $(t_K)$ is bounded above, and the condensed series converges by step 2.\n\n## Idea\n\nBetween consecutive powers of $2$, a decreasing sequence is squeezed between its values at the two ends. So the sum over the block from $2^k$ to $2^{k+1}$ is comparable to $2^k a_{2^k}$. The two series bound each other up to a factor $2$.\n",
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