cauchy condensation test

Cauchy condensation test

Let ${\sum }_{k=1}^{\infty }{a}_k$ be a series with $0\le a_{k+1}\le a_k\forall k$. Then

$$\sum _{k=1}^{\infty }{a}_k\text{ converges }⟺\sum _{k=0}^{\infty }{2}^k{a}_{2^k}\text{ converges}$$

The test is only applicable to non-negative and monotone null sequences.

Idea: Condense each block of $2^k$ terms into a single term, the block’s largest term $a_{2^k}$, repeated across the whole block:

$$\underset{2^k\text{ terms}}{\underbrace{a_{2^k}+a_{2^k+1}+\cdots +a_{2^{k+1}-1}}}⟶2^k{a}_{2^k}(\text{1 term})$$

Within this block, $a_n$ is squeezed between the endpoints (monotonicity):

$$a_{2^{k+1}}\le a_n\le a_{2^k}\text{for }{2}^k\le n<2^{k+1}$$

So the original series is squeezed between the condensed one and half of it:

$$2^k{a}_{2^{k+1}}\le \sum _{n=2^k}^{2^{k+1}-1}{a}_n\le 2^k{a}_{2^k}$$

Summed over $k$, this gives (… the 1/2 becomes visible):

$$\sum _{k=0}^{\infty }{2}^k{a}_{2^{k+1}}=\frac{1}{2}\sum _{j=1}^{\infty }{2}^j{a}_{2^j}\le \sum _{n=1}^{\infty }{a}_n\le \sum _{k=0}^{\infty }{2}^k{a}_{2^k}.$$

for all $k\in \mathbb{N}$. Original and condensed series converge or diverge together. $\square$

The harmonic series $a_n=1/n$

The p-series $a_n=1/n^2$