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Cauchy's condensation test says the following. Let $(a_n)$ be a decreasing sequence of nonnegative reals. Then the series $\sum_{n=1}^\infty a_n$ converges $\iff$ $\sum_{n=0}^\infty 2^na_{2^n}$ converges.

I've seen various proofs of this, but I haven't seen one that proves the $\Rightarrow$ direction using the contrapositive (show that if $\sum_{n=0}^\infty 2^na_{2^n}$ diverges, then $\sum_{n=1}^\infty a_n$ diverges). I want to do this by showing that a lower bound on the partial sums of $a_n$ is unbounded, but I don't know where to introduce the condensed series (which is a upper bound, not a lower bound, on the partial sums).

This problem is in Abbott's analysis text, and the hint is that the proof is similar to the proof that the harmonic series diverges (which shows that the partial sums are bounded below by an unbound sequence).

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Let $s_m = \sum_{k=1}^m a_k$. Consider the partial sums $t_m = \frac{1}{2} \sum_{k=1}^m 2^ka_{2^k}$. Notice that \begin{align*} s_{2^n} &= a_1 + a_2 + (a_3 + a_4) + (a_5 + a_6 + a_7 + a_8) + \ldots + (a_{2^{n-1}+1} + \ldots + a_{2^n}) \\ &\geq a_1 + a_2 + 2a_4 + 4a_8 + \ldots + 2^{n-1}a_{2^n} \\ &= a_1 + t_n \end{align*} If $\sum_{n=1}^\infty 2^na_{2^n}$ diverges, its partial sums must be unbounded, which implies that $(t_m)$ are also unbounded. So $(s_m)$ are unbounded, and the series diverges.

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