The Cauchy Condensation Test

For a decreasing sequence of positive terms, you can learn whether $\sum a_n$ converges by sampling it only at powers of $2$. This test is a neat application of comparison and settles the $p$-series without integrals.

Key idea

Cauchy condensation test. If $a_n \ge 0$ and $(a_n)$ is decreasing, then $\sum_{n=1}^{\infty} a_n$ converges if and only if $\sum_{k=0}^{\infty} 2^{k} a_{2^{k}}$ converges. Moreover $\sum a_n \le \sum 2^{k} a_{2^{k}}$.

Why it works

Group the terms in blocks of doubling length: $a_1$, then $a_2 + a_3$, then $a_4 + \cdots + a_7$, and so on; block $k$ is $a_{2^{k}} + \cdots + a_{2^{k+1} - 1}$, with $2^{k}$ terms. Because the terms decrease, each term of block $k$ is at most $a_{2^{k}}$ and at least $a_{2^{k+1}}$. So $$2^{k} a_{2^{k+1}} \le \text{block}_k \le 2^{k} a_{2^{k}}.$$ Summing the right inequality gives $\sum a_n \le \sum 2^{k} a_{2^{k}}$; summing the left one gives $\frac{1}{2}\sum_{k \ge 1} 2^{k} a_{2^{k}} \le \sum a_n$. So the two series converge or diverge together. $\blacksquare$

For the $p$-series, $2^{k} a_{2^{k}} = 2^{k}(2^{k})^{-p} = (2^{1 - p})^{k}$, a geometric series with ratio $2^{1 - p}$. It converges exactly when $2^{1 - p} < 1$, that is, $p > 1$. For $a_n = \frac{1}{n(\ln n)^{p}}$, $2^{k} a_{2^{k}} = \frac{1}{(k \ln 2)^{p}}$, a constant times a $p$-series in $k$: it converges exactly when $p > 1$. So $\sum \frac{1}{n \ln n}$ diverges, extremely slowly.

Why does sampling at powers of $2$ help? A decreasing sequence changes slowly, so all the terms between $2^{k}$ and $2^{k+1}$ are about the same size as $a_{2^{k}}$; there are $2^{k}$ of them, so the block is about $2^{k}a_{2^{k}}$. Condensing turns a series whose terms shrink like a power of $n$ into one whose terms shrink geometrically, or not at all, and geometric behavior is easy to judge. It also explains logarithms: substituting $n = 2^{k}$ turns $\ln n$ into $k \ln 2$, so one layer of logarithm becomes a power of $k$.

Worked examples

Worked example: The ratio of the condensed series

Apply condensation to $\sum \frac{1}{n^{3/2}}$.

$2^{k} a_{2^{k}} = \frac{2^{k}}{2^{3k/2}} = \left(2^{-1/2}\right)^{k}$.
The ratio is $2^{-1/2} = \frac{\sqrt{2}}{2} \approx 0.707 < 1$.

Answer: Geometric with ratio $\frac{\sqrt{2}}{2}$, so the series converges.

Worked example: A bound for a sum

Use $\sum a_n \le \sum_{k \ge 0} 2^{k} a_{2^{k}}$ to bound $\sum \frac{1}{n^{2}}$.

$2^{k} a_{2^{k}} = 2^{k} \cdot 2^{-2k} = \left(\frac{1}{2}\right)^{k}$.
$\sum_{k \ge 0} \left(\frac{1}{2}\right)^{k} = 2$.

Answer: $\sum \frac{1}{n^{2}} \le 2$ (the true value is $\frac{\pi^{2}}{6} \approx 1.645$).

Worked example: Logarithms in the denominator

Write the condensed term of $\sum_{n \ge 2} \frac{1}{n (\ln n)^{2}}$.

$a_{2^{k}} = \frac{1}{2^{k}(\ln 2^{k})^{2}} = \frac{1}{2^{k} k^{2} (\ln 2)^{2}}$.
Multiply by $2^{k}$: $\frac{1}{(\ln 2)^{2} k^{2}}$.

Answer: $\frac{1}{(\ln 2)^{2} k^{2}}$: a convergent $p$-series, so the original series converges.

Common mistake

Dropping the factor $2^{k}$, or ignoring "decreasing." The condensed series is $\sum 2^{k} a_{2^{k}}$, not $\sum a_{2^{k}}$; without the factor, $\sum \frac{1}{n}$ would seem to condense to the convergent $\sum \frac{1}{2^{k}}$. And the test needs decreasing terms: a sequence that is $0$ at the powers of $2$ and $\frac{1}{n}$ elsewhere has condensed series $0$ but diverges.

The Cauchy Condensation Test

Worked examples

Practice

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Sources and further reading