The Cauchy Condensation Test

Why Group Terms into Dyadic Blocks?

The Integral Test compares terms of a positive decreasing series with areas under a curve. Cauchy’s Condensation Test uses a different comparison: it groups consecutive terms into blocks whose lengths double. If the terms decrease, the terms in each block can be bounded by values at the block’s endpoints. The resulting estimates compare the original series with a much shorter series formed by sampling at powers of two.

For example, the block from $2^k$ through $2^{k+1}-1$ contains exactly $2^k$ terms. If each term in the block is close in size to the first one, then the block sum is close in size to $2^k a_{2^k}$. The condensation test makes this idea precise using inequalities; it does not require the terms to be close to one another in any stronger sense.

Definition: Given a positive sequence $(a_n)_{n=1}^{\infty}$, its condensed sequence is $c_k=2^k a_{2^k}$, for integers $k\geq0$. The associated condensed series is $\sum_{k=0}^{\infty}2^k a_{2^k}$.

The factor $2^k$ records the number of terms in the $k$-th dyadic block. The sequence values $a_{2^k}$ alone would omit this growing block size, so they would not in general capture the behavior of the original series.

The Comparison on One Block

Assume throughout that $a_n>0$ and $a_{n+1}\leq a_n$ for every $n$. For a fixed $k\geq0$, every index $n$ in the block $2^k\leq n\leq 2^{k+1}-1$ satisfies

a_{2^{k+1}}\leq a_n\leq a_{2^k}.

There are $2^k$ terms in the block, so adding these inequalities gives the following result.

Theorem (Dyadic Block Comparison): Suppose $(a_n)$ is positive and nonincreasing. Define $$ B_k=\sum_{n=2^k}^{2^{k+1}-1}a_n \quad\text{and}\quad c_k=2^k a_{2^k}. $$ Then, for every integer $k\geq0$, $$ \frac{c_{k+1}}{2}\leq B_k\leq c_k. $$

Proof. For $2^k\leq n\leq2^{k+1}-1$, nonincreasingness gives $a_{2^{k+1}}\leq a_n\leq a_{2^k}$. Summing across the $2^k$ indices yields

2^k a_{2^{k+1}}\leq \sum_{n=2^k}^{2^{k+1}-1}a_n \leq 2^k a_{2^k}.

The rightmost expression is $c_k$. Also,

2^k a_{2^{k+1}} =\frac{1}{2}\left(2^{k+1}a_{2^{k+1}}\right) =\frac{c_{k+1}}{2}.

Substituting these expressions proves both inequalities. $\square$

The two sides of the comparison have different roles. The upper bound says that a block sum is no greater than its condensed term. The lower bound says that the next condensed term is no greater than twice the block sum. Thus neither series can converge while the other diverges: the block sums and condensed terms control one another, up to a fixed factor and an index shift.

Cauchy’s Condensation Test

Theorem (Cauchy Condensation Test): Suppose $(a_n)_{n=1}^{\infty}$ is positive and nonincreasing. Then $$ \sum_{n=1}^{\infty}a_n \text{ converges} \quad\Longleftrightarrow\quad \sum_{k=0}^{\infty}2^k a_{2^k} \text{ converges}. $$

Proof. Let $B_k$ be the sum on the $k$-th dyadic block, as above. The blocks are disjoint and together contain every positive integer. Hence the partial sum through the end of the $m$-th block is

\sum_{n=1}^{2^{m+1}-1}a_n =\sum_{k=0}^{m}B_k.

All terms are positive, so the partial sums of the original series are increasing. They are bounded above if and only if the partial sums at these block endpoints are bounded above: one direction follows because endpoint partial sums are among the partial sums, and the other follows because any partial sum is no greater than the next endpoint partial sum. By the earlier theorem on bounded increasing partial sums, the original series converges exactly when $\sum_{k=0}^{\infty}B_k$ converges.

Now suppose the condensed series converges. The block comparison gives $B_k\leq c_k$ for each $k$, so the comparison principle for positive-term series shows that $\sum B_k$ converges. Therefore the original series converges.

Conversely, suppose the original series converges. Then the series of block sums $\sum B_k$ converges, since its partial sums are the original partial sums at the block endpoints. The lower bound in the block comparison gives

c_{k+1}\leq2B_k \quad (k\geq0).

Consequently, the partial sums of the condensed series are bounded: for every $m\geq1$,

\sum_{k=1}^{m}c_k \leq 2\sum_{k=0}^{m-1}B_k \leq 2\sum_{k=0}^{\infty}B_k.

The right side is finite and independent of $m$. Adding the single finite term $c_0=a_1$ preserves boundedness. Since the condensed terms are positive, bounded increasing partial sums imply that the condensed series converges. This proves the reverse direction and the test. $\square$

This proof uses positivity twice: it makes partial sums increase, and it permits convergence to be recognized from boundedness. Nonincreasingness supplies the block comparison. All three features matter; the test is not a general rule for sampling an arbitrary sequence at powers of two.

Worked Applications

Worked Example: A Rational Series

Consider

\sum_{n=1}^{\infty}\frac{1}{n^2+n}.

The terms are positive. The function $f(x)=1/(x^2+x)$ is decreasing for $x\geq1$, since

f'(x)=-\frac{2x+1}{(x^2+x)^2}<0.

Thus $a_n=f(n)$ is nonincreasing. Its condensed term is

2^k a_{2^k} =\frac{2^k}{(2^k)^2+2^k} =\frac{1}{2^k+1}.

For every $k\geq0$, $0<1/(2^k+1)\leq1/2^k$. The comparison geometric series $\sum_{k=0}^{\infty}1/2^k$ converges, so the condensed series converges. Cauchy’s Condensation Test proves that the original rational series converges.

Worked Example: The Logarithmic Harmonic Series

Define $a_1=1$ and $a_n=1/(n\log n)$ for $n\geq2$. The terms are positive. For $x\geq2$, the function $f(x)=1/(x\log x)$ has derivative

f'(x)=-\frac{\log x+1}{x^2(\log x)^2}<0.

Also $a_1=1>a_2=1/(2\log2)$, so the whole sequence is nonincreasing. For $k\geq1$, its condensed terms are

2^k a_{2^k} =\frac{2^k}{2^k\log(2^k)} =\frac{1}{k\log2}.

The sum of these terms diverges because it is the positive constant $1/\log2$ times the harmonic series. The initial condensed term $a_1$ does not affect divergence. By Cauchy’s Condensation Test, $\sum_{n=1}^{\infty}a_n$ diverges.

Worked Example: A Series with a Squared Logarithm

Define $a_1=2$ and $a_n=1/(n(\log n)^2)$ for $n\geq2$. The function $f(x)=1/(x(\log x)^2)$ is positive for $x\geq2$, and

f'(x)=-\frac{\log x+2}{x^2(\log x)^3}<0 \quad (x\geq2).

Moreover $a_1=2>a_2=1/(2(\log2)^2)$, so the sequence is nonincreasing. For $k\geq1$,

2^k a_{2^k} =\frac{2^k}{2^k(\log(2^k))^2} =\frac{1}{k^2(\log2)^2}.

The condensed series converges by comparison with the $p$-series $\sum_{k=1}^{\infty}1/k^2$. The extra term $a_1$ is finite, so it does not change convergence. Cauchy’s Condensation Test shows that the original series converges.

Why the Monotonicity Condition Cannot Be Dropped

The condition that the terms be nonincreasing is not merely a technical convenience. Without it, the sampled values at powers of two can miss most of the mass of a series. For instance, define $a_n=1/n$ when $n$ is not a power of two, and $a_n=0$ when $n$ is a power of two. This sequence is nonnegative but not nonincreasing. Every condensed term is zero, because $a_{2^k}=0$. The condensed series therefore converges. But the original series diverges: removing the terms $1/2^k$ from the harmonic series removes only a convergent geometric series, leaving a divergent sum. Thus the test’s conclusion would fail without monotonicity.

A useful way to apply the test is to check the hypotheses before simplifying the condensed terms. In many applications the formula is defined only from some integer onward, or is monotone only on a tail. In that situation, apply the test to the positive nonincreasing tail, using dyadic blocks from a suitable power of two. Adding or removing finitely many terms does not affect convergence, so this tail conclusion decides the convergence of the full series.

Check the terms.
Verify that the sequence is positive and nonincreasing, at least from some index onward.

Form the condensed terms.
Replace each term at index $2^k$ by $2^k a_{2^k}$, keeping track of any finite initial terms separately.

Test the condensed series.
Use a familiar comparison or convergence test on the simpler series.

Transfer the conclusion.
Cauchy’s Condensation Test gives the same convergence or divergence behavior for the original positive series.

Takeaway: For a positive nonincreasing sequence, the original series converges exactly when the series of dyadic block estimates $2^k a_{2^k}$ converges. The block comparison explains both the factor $2^k$ and the importance of monotonicity.

Check Your Understanding

Use the block comparison and hypotheses of Cauchy’s Condensation Test to answer these questions.

How many terms lie in the block from $2^k$ through $2^{k+1}-1$, and how does this produce the factor $2^k$?
Which inequality in the dyadic block comparison shows that convergence of the original series implies convergence of the condensed series?
Why is positivity needed when relating convergence to boundedness of partial sums?
For the series with terms $1/(n(\log n)^2)$, what is the condensed term for $k\geq1$?
How can sampling at powers of two miss divergent mass when the terms are not nonincreasing?

What You'll Learn