[proofplan] We establish the equivalence by bounding the partial sums $S_{2^n} = \sum_{j=1}^{2^n} a_j$ above and below in terms of the condensed partial sums $T_n = \sum_{k=0}^{n} 2^k a_{2^k}$. The key technique is grouping terms of the original series into dyadic blocks $[2^{k-1}+1, 2^k]$ and using the monotonicity of $(a_j)$ to bound each block above and below by a single term times the block length. The resulting two-sided inequality shows $S_{2^n}$ is bounded if and only if $T_n$ is bounded, and the [Monotone Convergence Theorem](/theorems/509) converts boundedness into convergence for these non-decreasing sequences. [/proofplan] [step:Bound each dyadic block using monotonicity of $(a_j)$] Since $(a_j)$ is decreasing and non-negative, for each $k \geq 1$ and each $j$ with $2^{k-1} \leq j \leq 2^k - 1$, we have $a_{2^k} \leq a_j \leq a_{2^{k-1}}$. Summing over the $2^{k-1}$ terms in the block $\{2^{k-1}, 2^{k-1}+1, \ldots, 2^k - 1\}$: \begin{align*} 2^{k-1} a_{2^k} \leq \sum_{j=2^{k-1}}^{2^k - 1} a_j \leq 2^{k-1} a_{2^{k-1}}. \end{align*} [guided] The idea behind the Cauchy Condensation Test is to group terms of the [series](/pages/1205) into blocks of geometrically increasing length and replace each block by a single representative term. Since $(a_j)$ is decreasing, all terms in a block are comparable to the endpoint values. For the block $\{2^{k-1}, 2^{k-1}+1, \ldots, 2^k - 1\}$ containing $2^{k-1}$ terms, monotonicity gives $a_{2^k} \leq a_j \leq a_{2^{k-1}}$ for each $j$ in this range. Summing over the block: \begin{align*} 2^{k-1} a_{2^k} \leq \sum_{j=2^{k-1}}^{2^k - 1} a_j \leq 2^{k-1} a_{2^{k-1}}. \end{align*} The lower bound uses the smallest term in the block ($a_{2^k}$) and the upper bound uses the largest ($a_{2^{k-1}}$). This two-sided estimate is what connects the original series to the condensed series. [/guided] [/step] [step:Establish upper and lower bounds relating $S_{2^n}$ and $T_n$] Let $S_{2^n} = \sum_{j=1}^{2^n} a_j$ and $T_n = \sum_{k=0}^{n} 2^k a_{2^k}$. For the upper bound, group terms starting from the second block: \begin{align*} S_{2^n} &= a_1 + \sum_{k=1}^{n} \sum_{j=2^{k-1}+1}^{2^k} a_j \\ &\leq a_1 + \sum_{k=1}^{n} 2^{k-1} a_{2^{k-1}} \\ &= a_1 + \sum_{k=0}^{n-1} 2^k a_{2^k} \\ &= a_1 + T_{n-1}. \end{align*} For the lower bound: \begin{align*} S_{2^n} &\geq \sum_{k=1}^{n} \sum_{j=2^{k-1}+1}^{2^k} a_j \\ &\geq \sum_{k=1}^{n} 2^{k-1} a_{2^k} \\ &= \frac{1}{2} \sum_{k=1}^{n} 2^k a_{2^k} \\ &= \frac{1}{2}(T_n - a_1). \end{align*} Combining: $\frac{1}{2}(T_n - a_1) \leq S_{2^n} \leq a_1 + T_{n-1}$. [/step] [step:Deduce the equivalence of convergence from the two-sided bound] Both $(S_N)$ and $(T_n)$ are non-decreasing sequences of non-negative terms. If $\sum_{j=1}^{\infty} a_j$ converges, then $(S_{2^n})$ is bounded above. The inequality $T_n \leq 2S_{2^n} + a_1$ shows $(T_n)$ is bounded above, so $\sum_{k=0}^{\infty} 2^k a_{2^k}$ converges. If $\sum_{k=0}^{\infty} 2^k a_{2^k}$ converges, then $(T_n)$ is bounded above. The inequality $S_{2^n} \leq a_1 + T_{n-1}$ shows $(S_{2^n})$ is bounded above. Since $(S_N)$ is non-decreasing (as $a_j \geq 0$) and bounded above along the subsequence $N = 2^n$, it is bounded above for all $N$. By the Monotone Convergence Theorem, $\sum_{j=1}^{\infty} a_j$ converges. [/step]