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Exercise 1473Calculus › Single-variable calculus › Series › Convergence testsM1Added 2 Nov 2026

Cauchy's condensation test

#condensation test#p-series#monotone terms

Statement

Let (un)(u_n) be a non-negative decreasing sequence. Prove that ∑un\sum u_n converges if and only if ∑2ku2k\sum2^ku_{2^k} converges. Deduce that ∑1np\sum\frac1{n^p} converges iff p>1p>1.

Hint

Group the terms of ∑un\sum u_n in blocks [2k,2k+1−1][2^k,2^{k+1}-1] and bound each block above and below using monotonicity.

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Solution

Block bounds. The block 2k≤n<2k+12^k\le n<2^{k+1} contains 2k2^k terms, each between u2k+1u_{2^{k+1}} and u2ku_{2^k}:
2ku2k+1≤∑n=2k2k+1−1un≤2ku2k.2^ku_{2^{k+1}}\le\sum_{n=2^k}^{2^{k+1}-1}u_n\le2^ku_{2^k}.

If the condensed series converges, summing the right inequality over kk shows ∑un\sum u_n is bounded above by ∑2ku2k<∞\sum2^ku_{2^k}<\infty, so it converges (positive terms).

If the condensed series diverges, the left inequality gives ∑nun≥∑k2ku2k+1=12∑k≥12ku2k\sum_n u_n\ge\sum_k2^ku_{2^{k+1}}=\frac12\sum_{k\ge1}2^ku_{2^k}, which diverges.

Application. For un=1npu_n=\frac1{n^p} with p>0p>0 (decreasing): 2ku2k=2k(1−p)2^ku_{2^k}=2^{k(1-p)}, a geometric series that converges iff 21−p<12^{1-p}<1, i.e. p>1p>1. For p≤0p\le0 the terms do not tend to 00.