Proof of Cauchy Condensation

A Proof at Every Partial-Sum Cutoff

Cauchy’s Condensation Test was established in the previous tutorial by grouping terms into dyadic blocks. Here we refine that proof: instead of comparing only sums that end at block boundaries, we bound an arbitrary partial sum using the condensed terms. The resulting estimates give a direct proof of the test and also quantify how the two series control one another.

Throughout, let $(a_n)_{n=1}^{\infty}$ be positive and nonincreasing, and write

c_k=2^k a_{2^k}, \qquad C_m=\sum_{k=0}^{m}c_k, \qquad S_N=\sum_{n=1}^{N}a_n.

The Dyadic Block Comparison from the previous tutorial says that if

B_k=\sum_{n=2^k}^{2^{k+1}-1}a_n,

then $c_{k+1}/2\leq B_k\leq c_k$. We will use that comparison rather than derive it again.

Bounds for Arbitrary Partial Sums

Theorem (Finite-Cutoff Condensation Bounds): Suppose $(a_n)$ is positive and nonincreasing, and define $S_N$, $c_k$, and $C_m$ as above. If $2^m\leq N<2^{m+1}$, then $$ \frac{C_m-c_0}{2}\leq S_N\leq C_m. $$ Equivalently, $C_m\leq 2S_N+c_0$.

Proof. The blocks with indices $k=0,\ldots,m-1$ are complete in $S_N$, because their last indices are $2^{k+1}-1\leq2^m-1\leq N$. The remaining terms in $S_N$, if any, have indices from $2^m$ through $N$.

For the upper bound, each complete block satisfies $B_k\leq c_k$ by the Dyadic Block Comparison. In the remaining part, there are at most $2^m$ terms. Since the sequence is nonincreasing, every one of them is at most $a_{2^m}$. Therefore

S_N \leq \sum_{k=0}^{m-1}c_k+2^m a_{2^m} =\sum_{k=0}^{m}c_k =C_m.

When $m=0$, the sum over complete blocks is empty, and the same estimate follows from $N<2$, so $N=1$ and $S_1=a_1=c_0=C_0$.

For the lower bound, keep only the complete blocks. The lower estimate $B_k\geq c_{k+1}/2$ gives

S_N\geq\sum_{k=0}^{m-1}B_k \geq\frac{1}{2}\sum_{k=0}^{m-1}c_{k+1} =\frac{1}{2}\sum_{j=1}^{m}c_j =\frac{C_m-c_0}{2}.

If $m=0$, this last lower bound is $0\leq S_N$, which holds because the terms are positive. Rearranging the lower bound proves $C_m\leq2S_N+c_0$. $\square$

The estimates work for every $N$ in the indicated dyadic range. The upper estimate accounts for the incomplete block by bounding all its terms by its first term; the lower estimate needs no information about that incomplete block, since all its terms are positive. This difference explains why the two sides have different forms.

Recovering the Condensation Equivalence

The finite-cutoff bounds give a proof of the Cauchy Condensation Test using only partial sums. If $\sum_{k=0}^{\infty}c_k$ converges, then its partial sums $C_m$ are bounded above. Given any $N$, choose the unique integer $m\geq0$ such that $2^m\leq N<2^{m+1}$. The upper estimate gives $S_N\leq C_m$, so the partial sums $S_N$ are bounded above. They are increasing because $a_n>0$; by the earlier theorem on bounded increasing partial sums, $\sum_{n=1}^{\infty}a_n$ converges.

Conversely, suppose $\sum_{n=1}^{\infty}a_n$ converges. Its partial sums are bounded above, say $S_N\leq M$ for every $N$. For each $m$, the index $N=2^m$ lies in the range $2^m\leq N<2^{m+1}$. Thus the finite-cutoff estimate gives

C_m\leq 2S_{2^m}+c_0\leq 2M+c_0.

The condensed partial sums $C_m$ are increasing because $c_k>0$, and the displayed inequality bounds them above. The same theorem on bounded increasing partial sums shows that $\sum_{k=0}^{\infty}c_k$ converges. Since $c_k=2^k a_{2^k}$, this proves the convergence equivalence in Cauchy’s Condensation Test. The test itself is not a new conclusion here; the finite-cutoff estimates supply a sharper way to see why it holds.

A Quantitative Comparison of Tails

The block estimates also compare a finite collection of consecutive dyadic blocks. This is useful when studying remainders: it locates the original-series tail and the condensed-series tail on the same range of block indices.

Theorem (Finite Dyadic Tail Comparison): For integers $0\leq r\leq s$, $$ \frac{1}{2}\sum_{j=r+1}^{s+1}c_j \leq \sum_{n=2^r}^{2^{s+1}-1}a_n \leq \sum_{k=r}^{s}c_k. $$

Proof. The interval of indices from $2^r$ through $2^{s+1}-1$ is the disjoint union of the blocks from $2^k$ through $2^{k+1}-1$, for $k=r,\ldots,s$. Consequently,

\sum_{n=2^r}^{2^{s+1}-1}a_n =\sum_{k=r}^{s}B_k.

Applying the Dyadic Block Comparison to each block and adding the lower bounds gives

\sum_{k=r}^{s}B_k \geq\frac{1}{2}\sum_{k=r}^{s}c_{k+1} =\frac{1}{2}\sum_{j=r+1}^{s+1}c_j.

Adding the upper bounds $B_k\leq c_k$ instead gives $\sum_{k=r}^{s}B_k\leq\sum_{k=r}^{s}c_k$. Together these inequalities prove the result. $\square$

The shift by one index in the lower bound matters. A block beginning at $2^k$ controls $c_{k+1}$ from below, not $c_k$. Keeping that shift visible prevents a common indexing error when applying condensation to tails or remainders.

Worked Applications of the Bounds

Worked Example: A Simple Divergent Sequence

Let $a_n=1/(n+2)$ for $n\geq1$. These terms are positive and decrease as $n$ increases. The condensed terms are

c_k=2^k a_{2^k} =\frac{2^k}{2^k+2}.

Since $2^k\geq1$, we have $2^k+2\leq3\cdot2^k$, and hence $c_k\geq1/3$. Therefore $C_m\geq(m+1)/3$, which is unbounded as $m$ increases. The finite-cutoff theorem shows that convergence of the original series would force the $C_m$ to be bounded. Thus $\sum_{n=1}^{\infty}1/(n+2)$ diverges.

Worked Example: A Convergent Logarithmic Series

For $n\geq1$, define

a_n=\frac{1}{n(1+\log n)^{3/2}}.

For $x\geq1$, the function $f(x)=1/[x(1+\log x)^{3/2}]$ is positive and has derivative

f'(x) =-\frac{(1+\log x)+3/2}{x^2(1+\log x)^{5/2}}<0.

Thus the sequence meets the hypotheses. Its condensed terms are

c_k =2^k a_{2^k} =\frac{1}{(1+k\log2)^{3/2}}.

For $k\geq1$, $1+k\log2\geq k\log2$, so

0<c_k\leq\frac{1}{(\log2)^{3/2}k^{3/2}}.

The comparison series is a constant multiple of a convergent $p$-series. The term $c_0=1$ is finite, so the condensed series converges. By the finite-cutoff proof of Cauchy’s Condensation Test, the original series converges.

Worked Example: A Divergent Square-Root Logarithmic Series

Now define $a_n=1/[n\sqrt{1+\log n}]$ for $n\geq1$. The associated function is positive on $[1,\infty)$, and its derivative is

\frac{d}{dx}\left(\frac{1}{x\sqrt{1+\log x}}\right) =-\frac{(1+\log x)+1/2}{x^2(1+\log x)^{3/2}}<0.

The sequence is therefore positive and decreasing. Its condensed terms satisfy

c_k=\frac{1}{\sqrt{1+k\log2}}.

For $k\geq1$, $1+k\log2\leq(1+\log2)k$. Taking reciprocals and square roots yields

c_k\geq\frac{1}{\sqrt{1+\log2}}\frac{1}{\sqrt{k}}.

The comparison $p$-series with exponent $1/2$ diverges, so the condensed partial sums are unbounded. The finite-cutoff bounds rule out convergence of the original series, which therefore diverges.

What the Proof Depends On

The proof uses nonincreasingness to control the unfinished block in the upper estimate and to obtain the Dyadic Block Comparison. Without monotonicity, a block could have small values at its first index and much larger values later, so its condensed term would not necessarily bound its block sum. Positivity has a different role: it makes partial sums increasing, allowing boundedness to imply convergence. These hypotheses are not interchangeable, and both must be checked before applying the test.

The finite-cutoff bounds are also useful as a diagnostic. If a calculation suggests that $C_m$ grows without bound, then $S_{2^m}$ must grow without bound as well, since $C_m\leq2S_{2^m}+c_0$. If instead the $C_m$ stay bounded, then every $S_N$ stays bounded, including partial sums that stop in the middle of a dyadic block. That is the central advantage of proving estimates for arbitrary cutoffs.

Takeaway: For a positive nonincreasing sequence, every partial sum between two consecutive powers of two is bounded above by the corresponding condensed partial sum, while the condensed partial sum is bounded above by twice the original partial sum plus $a_1$. These finite estimates yield Cauchy condensation and sharpen its use for tails.

Check Your Understanding

Use the finite-cutoff and dyadic-tail comparisons to answer the following questions.

If $2^m\leq N<2^{m+1}$, which part of the proof gives the upper bound $S_N\leq C_m$?
Why does the lower estimate for $S_N$ involve $C_m-c_0$, rather than all of $C_m$?
How does boundedness of the condensed partial sums imply boundedness of every original partial sum?
In the finite dyadic tail comparison, why does the lower bound involve $c_{r+1},\ldots,c_{s+1}$?
For $a_n=1/[n\sqrt{1+\log n}]$, what lower bound on the condensed terms proves divergence?

What You'll Learn