Cauchy Condensation Test

Block $n$ holds $2^n$ terms in $[2^n, 2^{n+1})$.

Since the sequence decreases, each term is at most $a_{2^n}$ and at least $a_{2^{n+1}}$.

Hence the block sum $B_n = \sum_{k=2^n}^{2^{n+1}-1} a_k$ satisfies $\frac{1}{2}C_{n+1} = 2^n a_{2^{n+1}} \le B_n \le 2^n a_{2^n} = C_n$ (with $C_n := 2^n a_{2^n}$).

$(\Leftarrow)$ Summing $B_j \le C_j$ for $j = 0$ to $N$ gives $S_{2^{N+1}-1} = \sum_{j=0}^{N} B_j \le \sum_{j=0}^{N} C_j = T_N$.

If $\sum C_j$ converges, then $T_N \le T := \sum_{j=0}^{\infty} C_j < \infty$.

Extend from binary partial sums to all partial sums: for arbitrary $m$, choose $N$ with $m \le 2^{N+1} - 1$; since $a_k \ge 0$, $S_m$ is increasing, so $S_m \le S_{2^{N+1}-1} \le T_N \le T$.

Hence all partial sums of $\sum a_n$ are bounded and converge.

$(\Rightarrow)$ Conversely, summing $C_{j+1} \le 2 B_j$ for $j = 0$ to $N$ gives $T_{N+1} - C_0 = \sum_{j=1}^{N+1} C_j \le 2 \sum_{j=0}^{N} B_j = 2 S_{2^{N+1}-1}$. If $\sum a_n$ converges, the right side is bounded, so the partial sums of $\sum C_j$ are bounded and converge.

Numerical check (harmonic series $a_n = 1/n$, block $n = 2$): $B_2 = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \approx 0.7595$, $C_2 = 4 \cdot \frac{1}{4} = 1$, $\frac{1}{2}C_3 = \frac{1}{2} \cdot 8 \cdot \frac{1}{8} = 0.5$ — matching the caption's sandwich $0.5 \le 0.7595 \le 1$.

For $p > 0$, $a_n = \frac{1}{n^p}$ is a decreasing sequence of positive numbers. The condensed terms are $C_n = 2^n \cdot \frac{1}{(2^n)^p} = 2^n \cdot 2^{-np} = 2^{n(1-p)} = (2^{1-p})^n$, a geometric series with ratio $r = 2^{1-p}$.

A geometric series converges for $r < 1$ and diverges for $r \ge 1$. And $r = 2^{1-p} < 1$ is equivalent to $1 - p < 0$, i.e. $p > 1$ (for $p = 1$, $r = 1$ gives the harmonic series).

By the condensation test, $\sum \frac{1}{n^p}$ shares its fate with the condensed series, so it converges for $p > 1$ and diverges for $p \le 1$ (for $p \le 0$ the terms do not even tend to $0$).

Numerical check ($p = 2$, block $n = 2$): $B_2 = \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} \approx 0.1507$, $C_2 = 4 \cdot \frac{1}{16} = 0.25$, $\frac{1}{2}C_3 = \frac{1}{2} \cdot 8 \cdot \frac{1}{64} = 0.0625$ — matching the caption's $0.0625 \le 0.1507 \le 0.25$, and the condensed terms are exactly $2^{-n}$.

First verify the decreasing hypothesis: for $x \ge 2$, $x \ln x$ has derivative $\ln x + 1 > 0$, so $x \ln x$ is increasing and its reciprocal $a_n = \frac{1}{n \ln n}$ is decreasing and positive for $n \ge 2$.

Reindexing is also needed: the condensation in problem 1 starts at $n = 1$ and includes a $C_0$ term, but this series starts at $n = 2$ and $C_0 = \frac{1}{0 \cdot \ln 2}$ is singular.

So start the condensed sum at $n \ge 1$ ($C_n = 2^n a_{2^n}$ with $2^n \ge 2$ is well defined; finitely many terms do not affect convergence).

Condensing with $\ln 2^n = n \ln 2$ gives $C_n = 2^n \cdot \frac{1}{2^n \cdot n \ln 2} = \frac{1}{\ln 2}\cdot\frac{1}{n}$ for $n \ge 1$, so the condensed series $\sum_{n \ge 1} C_n$ is $\frac{1}{\ln 2}$ times the harmonic series — divergent.

Hence $\sum_{n=2}^{\infty} \frac{1}{n \ln n}$ diverges too.

But $b_n = \frac{1}{n (\ln n)^2}$ also needs the decreasing verification: for $x \ge 2$, $x (\ln x)^2$ has derivative $(\ln x)^2 + 2\ln x = \ln x (\ln x + 2) > 0$, so $n (\ln n)^2$ is increasing and its reciprocal $b_n$ is decreasing.

With the same reindexing (condense from $n \ge 1$), $b_n$ condenses to $C_n = 2^n \cdot \frac{1}{2^n (n \ln 2)^2} = \frac{1}{(\ln 2)^2}\cdot\frac{1}{n^2}$ for $n \ge 1$.

Since $\frac{1}{(\ln 2)^2} \approx 2.0814$ is a constant, the condensed series is a constant multiple of the convergent $\sum_{n \ge 1} \frac{1}{n^2}$ — indeed summing to $\frac{1}{(\ln 2)^2}\cdot\frac{\pi^2}{6} \approx 2.0814 \cdot 1.6449 \approx 3.4237$.

Hence $\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}$ converges.

Numerical check (first series, block $n = 2$): $B_2 = \sum_{k=4}^{7} \frac{1}{k \ln k} \approx 0.4710$, $C_2 = \frac{1}{2\ln 2} \approx 0.7213$, $\frac{1}{2}C_3 = \frac{1}{6\ln 2} \approx 0.2404$ — matching the caption's $0.2404 \le 0.4710 \le 0.7213$. One extra power of $\ln$ is exactly the borderline between divergence and convergence.