Cauchy Condensation Test

Group the terms ana_n into blocks [2n,2n+1)[2^n, 2^{n+1}): each block sum is sandwiched between condensed terms 2na2n2^n a_{2^n}, so the original and condensed series share their fate.

2na2n+1≤∑k=2n2n+1−1ak≤2na2n2^n a_{2^{n+1}} \le \sum_{k=2^n}^{2^{n+1}-1} a_k \le 2^n a_{2^n}
∑n≥11np converges  ⟺  p>1\sum_{n\ge 1}\frac{1}{n^p}\ \text{converges} \iff p>1

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Examples & Exercises

Let an≥0a_n \ge 0 be decreasing. Prove that ∑n=1∞an\sum_{n=1}^{\infty} a_n converges if and only if ∑n=0∞2na2n\sum_{n=0}^{\infty} 2^n a_{2^n} converges (the Cauchy condensation test).

2na2n+1≤∑k=2n2n+1−1ak≤2na2n,Cn:=2na2n2^n a_{2^{n+1}} \le \sum_{k=2^n}^{2^{n+1}-1} a_k \le 2^n a_{2^n}, \qquad C_n := 2^n a_{2^n}
Block n=2n=2 [4,8)[4, 8): 0.5=12C3≤B2≈0.7595≤C2=10.5 = \frac{1}{2}C_3 \le B_2 \approx 0.7595 \le C_2 = 1 — the block sum sits between the condensed terms.
Show proof
  1. Block nn holds 2n2^n terms in [2n,2n+1)[2^n, 2^{n+1}). Since the sequence decreases, each term is at most a2na_{2^n} and at least a2n+1a_{2^{n+1}}. Hence the block sum Bn=∑k=2n2n+1−1akB_n = \sum_{k=2^n}^{2^{n+1}-1} a_k satisfies 12Cn+1=2na2n+1≤Bn≤2na2n=Cn\frac{1}{2}C_{n+1} = 2^n a_{2^{n+1}} \le B_n \le 2^n a_{2^n} = C_n (with Cn:=2na2nC_n := 2^n a_{2^n}).
  2. Summing Bj≤CjB_j \le C_j for j=0j = 0 to NN gives S2N+1−1=∑j=0NBj≤∑j=0NCj=TNS_{2^{N+1}-1} = \sum_{j=0}^{N} B_j \le \sum_{j=0}^{N} C_j = T_N. If ∑Cj\sum C_j converges, then TNT_N is bounded, so the partial sums of ∑an\sum a_n (with ak≥0a_k \ge 0) are bounded and converge.
  3. Conversely, summing Cj+1≤2BjC_{j+1} \le 2 B_j for j=0j = 0 to NN gives TN+1−C0=∑j=1N+1Cj≤2∑j=0NBj=2S2N+1−1T_{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 ∑an\sum a_n converges, the right side is bounded, so the partial sums of ∑Cj\sum C_j are bounded and converge.
  4. Numerical check (harmonic series an=1/na_n = 1/n, block n=2n = 2): B2=14+15+16+17≈0.7595B_2 = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} \approx 0.7595, C2=4⋅14=1C_2 = 4 \cdot \frac{1}{4} = 1, 12C3=12⋅8⋅18=0.5\frac{1}{2}C_3 = \frac{1}{2} \cdot 8 \cdot \frac{1}{8} = 0.5 — matching the caption's sandwich 0.5≤0.7595≤10.5 \le 0.7595 \le 1.

Use the condensation test to show that the pp-series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges for p>1p > 1 and diverges for p≤1p \le 1.

2n⋅1(2n)p=2n(1−p)=(21−p)n(geometric series)2^n \cdot \frac{1}{(2^n)^p} = 2^{n(1-p)} = (2^{1-p})^n \quad \text{(geometric series)}
p=2p = 2, block [4,8)[4, 8): 0.0625=12C3≤B2≈0.1507≤C2=0.250.0625 = \frac{1}{2}C_3 \le B_2 \approx 0.1507 \le C_2 = 0.25 — the condensed terms are geometric with ratio 1/21/2.
Show proof
  1. For p>0p > 0, an=1npa_n = \frac{1}{n^p} is a decreasing sequence of positive numbers. The condensed terms are Cn=2n⋅1(2n)p=2n⋅2−np=2n(1−p)=(21−p)nC_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=21−pr = 2^{1-p}.
  2. A geometric series converges for r<1r < 1 and diverges for r≥1r \ge 1. And r=21−p<1r = 2^{1-p} < 1 is equivalent to 1−p<01 - p < 0, i.e. p>1p > 1 (for p=1p = 1, r=1r = 1 gives the harmonic series).
  3. By the condensation test, ∑1np\sum \frac{1}{n^p} shares its fate with the condensed series, so it converges for p>1p > 1 and diverges for p≤1p \le 1 (for p≤0p \le 0 the terms do not even tend to 00).
  4. Numerical check (p=2p = 2, block n=2n = 2): B2=116+125+136+149≈0.1507B_2 = \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} \approx 0.1507, C2=4⋅116=0.25C_2 = 4 \cdot \frac{1}{16} = 0.25, 12C3=12⋅8⋅164=0.0625\frac{1}{2}C_3 = \frac{1}{2} \cdot 8 \cdot \frac{1}{64} = 0.0625 — matching the caption's 0.0625≤0.1507≤0.250.0625 \le 0.1507 \le 0.25, and the condensed terms are exactly 2−n2^{-n}.

Show that ∑n=2∞1nln⁡n\sum_{n=2}^{\infty} \frac{1}{n \ln n} diverges while ∑n=2∞1n(ln⁡n)2\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2} converges, using the condensation test.

2n⋅12nln⁡2n=1nln⁡2=1ln⁡2⋅1n2^n \cdot \frac{1}{2^n \ln 2^n} = \frac{1}{n \ln 2} = \frac{1}{\ln 2}\cdot\frac{1}{n}
2n⋅12n(ln⁡2n)2=1(ln⁡2)2⋅1n22^n \cdot \frac{1}{2^n (\ln 2^n)^2} = \frac{1}{(\ln 2)^2}\cdot\frac{1}{n^2}
Block [4,8)[4, 8): 0.2404=12C3≤B2≈0.4710≤C2=12ln⁡2≈0.72130.2404 = \frac{1}{2}C_3 \le B_2 \approx 0.4710 \le C_2 = \frac{1}{2\ln 2} \approx 0.7213 — the condensed series is a constant multiple of the harmonic series, hence divergent.
Show proof
  1. an=1nln⁡na_n = \frac{1}{n \ln n} is decreasing and positive for n≥2n \ge 2. Condensing with ln⁡2n=nln⁡2\ln 2^n = n \ln 2 gives Cn=2n⋅12n⋅nln⁡2=1ln⁡2⋅1nC_n = 2^n \cdot \frac{1}{2^n \cdot n \ln 2} = \frac{1}{\ln 2}\cdot\frac{1}{n}, so the condensed series is 1ln⁡2\frac{1}{\ln 2} times the harmonic series — divergent. Hence ∑n=2∞1nln⁡n\sum_{n=2}^{\infty} \frac{1}{n \ln n} diverges too.
  2. But bn=1n(ln⁡n)2b_n = \frac{1}{n (\ln n)^2} condenses to Cn=2n⋅12n(nln⁡2)2=1(ln⁡2)2⋅1n2C_n = 2^n \cdot \frac{1}{2^n (n \ln 2)^2} = \frac{1}{(\ln 2)^2}\cdot\frac{1}{n^2}. Since 1(ln⁡2)2≈2.0814\frac{1}{(\ln 2)^2} \approx 2.0814 is a constant, the condensed series is a constant multiple of the convergent ∑1n2\sum \frac{1}{n^2} — indeed summing to 1(ln⁡2)2⋅π26≈2.0814⋅1.6449≈3.4237\frac{1}{(\ln 2)^2}\cdot\frac{\pi^2}{6} \approx 2.0814 \cdot 1.6449 \approx 3.4237. Hence ∑n=2∞1n(ln⁡n)2\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2} converges.
  3. Numerical check (first series, block n=2n = 2): B2=∑k=471kln⁡k≈0.4710B_2 = \sum_{k=4}^{7} \frac{1}{k \ln k} \approx 0.4710, C2=12ln⁡2≈0.7213C_2 = \frac{1}{2\ln 2} \approx 0.7213, 12C3=16ln⁡2≈0.2404\frac{1}{2}C_3 = \frac{1}{6\ln 2} \approx 0.2404 — matching the caption's 0.2404≤0.4710≤0.72130.2404 \le 0.4710 \le 0.7213. One extra power of ln⁡\ln is exactly the borderline between divergence and convergence.

6. Series