(2.4.6) [Cauchy Condensation Test] Suppose $(b_n)$ is decreasing and satisfies $b_n \geq 0$ for all $n \in \mathbf{N}$. Then, the series $\sum_{n=1}^{\infty} b_n$ converges if and only if the series $$ \begin{align*} \sum_{n=1}^{\infty} 2^nb_{2^n} = b_1 + 2b_2 + 4b_4 + 8b_8 + 16b_{16} + ... \end{align*} $$

converges.

Proof:

Let $(b_n)$ be a sequence such that $b_n \geq 0$ for all $n \in \mathbf{N}$. For the backward direction, suppose $\sum_{n=1}^{\infty} 2^nb_{2^n}$ converges. Let $(t_m)$ be the sequence of partial sums with $t_m$ defined as

$$ \begin{align*} t_m = b_1 + 2b_2 + 4b_4 + 8b_8 + ... + 2^mb_{2^m}. \end{align*} $$

Since $\sum_{n=1}^{\infty} 2^nb_{2^n}$ converges, then we know that $(t_m)$ converges. By theorem 2.3.2 we know that every convergent sequence is bounded and so $(t_m)$ must be bounded as well. So there exists some $M > 0$ such that $|t_m| \leq M$ for all $m \in \mathbf{N}$. But since $t_m \geq 0$, then we can just write $t_m \leq M$.

Now, to prove that $\sum_{n=1}^{\infty} b_n$ converges, we need to show that the sequence of partial sums converges. Let $(s_k)$ be the sequence of partial sums with $s_k$ defined as

$$ \begin{align*} s_k = b_1 + b_2 + b_3 + b_4 + ... + b_k. \end{align*} $$

The goal is to show that it’s bounded and monotone and then using the Montone Convergence Theorem to conclude that it converges. We are given that $b_n \geq 0$ for all $n \in \mathbf{N}$. Therefore, we know that $(s_k)$ is increasing and therefore monotonic. To see that it’s bounded, fix $k$ and let $m$ be large enough to ensure that $k \leq 2^{m+1} - 1$. We do so because we want to ensure the inequality,

$$ \begin{align*} s_k \leq s_{2^{m+1} - 1}, \end{align*} $$

holds. And we do this because we want to derive a bound on $s_k$ from expanding $s_{2^{m+1} - 1}$. To see this, expand $s_{2^{m+1} - 1}$ such that

$$ \begin{align*} s_{2^{m+1}-1} &= b_1 + (b_2 + b_3) + (b_4 + b_5 + b_6 + b_7) ... + (b_{2^m} + ... + b_{2^{m+1}-1}) \\ &\leq b_1 + (b_2 + b_2) + (b_4 + b_4 + b_4 + b_4) ... + (b_{2^m} + ... + b_{2^m}) \quad \text{(since $(b_n)$ is decreasing)}\\ &= b_1 + 2b_2 + 4b_4 ... + 2^mb_{2^m}\\ & = t_m \leq M. \end{align*} $$

From this we see that $s_k \leq t_k \leq M$, and the sequence $(s_k)$ is bounded. Therefore, by the Montone Convergence Theorem, we can conclude that $\sum_{n=1}^{\infty} b_n$ converges.

For the forward direction, we’ll prove the contrapositive of the statement and so suppose that $\sum_{n=1}^{\infty} 2^nb_{2^n}$ diverges, we will prove that $\sum_{n=1}^{\infty} b_n$ diverges as well. Because $\sum_{n=1}^{\infty} 2^nb_{2^n}$ diverges, then the sequence of partial sums $(t_m)$ also diverges. But we also know that $(t_m)$ is increasing and that $b_n \geq 0$ for all $n \in \mathbf{N}$. So this means that $(t_m)$ must be unbounded (If it was bounded, then it will converge and we assumed that it doesn’t).

Consider now the series $\sum_{n=1}^{\infty} b_n$ and its sequence of partial sums $(s_k)$. Fix $k$ and pick $m$ such that $2^m \leq k \leq 2^{m+1}$. From this, we can see that $s_k \geq s_{2^m}$ since $(s_k)$ is an increasing sequence. Expand $s_{2^m}$ to see that

$$ \begin{align*} s_{2^m} &= b_1 + b_2 + (b_3 + b_4) + (b_5 + b_6 + b_7 + b_8) + ... + (b_{2^{m-1}} + ... + b_{2^m}) \\ &\geq b_1 + b_2 + (b_4 + b_4) + (b_8 + b_8 + b_8 + b_8) + ... + (b_{2^{m}} + ... + b_{2^m}) \\ &= b_1 + b_2 + 2b_4 + 4b_8 + ... + 2^{m-1}b_{2^m} \\ &= \frac{1}{2}b_1 + \frac{1}{2}(2b_2 + 4b_4 + 8b_8 + ... + 2^{m}b_{2^m}) \\ &= \frac{1}{2}b_1 + \frac{1}{2}t_m. \\ \end{align*} $$

From this we see that $s_k \geq s_{2^m} \geq \frac{1}{2}b_1 + \frac{1}{2}t_m$. But we already concluded $(t_m)$ is unbounded and therefore, $(s_k)$ is unbounded. By the Montone Convergence Theorem, we know a sequence converges if it’s bounded and monotone. Since we $(s_k)$ is monotone but unbounded then it diverges as we wanted to show. $\blacksquare$

Other Definitions and Properties

Fo the absolute value function definition and other properties, see here.
For the definitions of sequences and what it means to for a sequence to converge, see this, for subsequences see this.
For the definitions of series, and what it means to for a series to converge, see this.
For the "show the limit" template and an example, see this.

References: