# Cauchy Condensation Test

## Contents

## Theorem

Let $\left \langle {a_n} \right \rangle: n \mapsto a\left({n}\right)$ be a decreasing sequence of strictly positive terms in $\R$ which converges with a limit of zero.

That is, for every $n$ in the domain of $\left \langle {a_n} \right \rangle$: $a_n > 0$, $a_n \ge a_{n+1}$, and $a_n \to 0$ as $n \to +\infty$.

Then the series:

- $\displaystyle \sum_{n \mathop = 1}^\infty a_n$

converges if and only if the condensed series:

- $\displaystyle \sum_{n \mathop = 1}^\infty 2^n a\left({2^n}\right)$

converges.

## Proof

### Necessary Condition

We will first show that if the condensed series $\displaystyle \sum_{n \mathop = 1}^\infty 2^n a\left({2^n}\right)$ converges, then $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges as well.

Assume $\displaystyle \sum_{n \mathop = 1}^\infty 2^n a \left({2^n}\right)$ converges.

Consider the graph of $\langle {a_n}\rangle$ and the partial sums of $\sum 2^n a \left({2^n}\right)$:

The dotted black line represents the sequence $\langle {a_n}\rangle$.

The $n$th rectangle has:

- Base equal to $2^n$

- Altitude equal to $a\left({2^n}\right)$

The series of partial sums of the areas of the rectangles are:

- $\displaystyle \sum_{n \mathop = 1}^N 2^n a \left({2^n}\right)$

which is precisely the defined condensed series.

The series $\sum a_n$ can be viewed as the sum of the areas of rectangles with width $1$ and height $a_n$.

The diagram suggests that the partial sums of $\sum a_n$ are not greater than the condensed partial sums:

- $\displaystyle \sum_{n \mathop = 1}^N a_n \le \sum_{n \mathop = 1}^N 2^n a \left({2^n}\right)$

To formalize this claim, observe that because $\langle {a_n}\rangle$ is decreasing:

- $a_1 + \underbrace{a_2 + a_3}_{\le 2a_2} + \underbrace{a_4 + a_5 + a_6 + a_7}_{\le 4a_4} + \cdots + a_N \le a_1 + 2a_2 + 4a_4 + \cdots + 2^N a\left({2^N}\right)$

As $n \to +\infty$, the right hand side converges by hypothesis.

Hence $\displaystyle \sum_{n \mathop = 1}^{\infty} a_n$ also converges, by the Comparison Test.

$\Box$

### Sufficient Condition

We will show that if $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges, the condensed series $\displaystyle \sum_{n \mathop = 1}^\infty 2^n a\left({2^n}\right)$ converges as well.

Assume $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ converges.

From the Combination Theorem for Sequences, $2\displaystyle \sum_{n \mathop = 1}^\infty a_n = \displaystyle \sum_{n \mathop = 1}^\infty 2a_n$ converges as well.

Consider, then, the graph of $\langle {a_n}\rangle$ and of the partial sums of what will be shown to equal $\displaystyle \frac 1 2 \sum 2^n a \left({2^n}\right)$:

The dotted black line represents the sequence $\langle {a_n}\rangle$.

The $n$th rectangle has:

- Base equal to $2^n$

- Altitude equal to $a\left( {2 \cdot 2^n }\right)$

The series of partial sums of the areas of the rectangles are:

- $\displaystyle \sum_{n \mathop = 1}^N 2^n a \left(2 \cdot {2^n}\right)$

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n \mathop = 1}^N 2^n a \left({ 2^{n+1} }\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{n-1 \mathop = 1}^N 2^{n-1} a \left({ 2^n }\right)\) | replacing $n$ with $n-1$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac 1 2 \sum_{n \mathop = 2}^N 2^n a \left({ 2^n }\right)\) |

From Deletion of Terms from a Sequence, it is justified to examine only the behavior of $n$ for $n \ge 2$.

The series $\sum a_n$ can be viewed as the sum of the area of rectangles with height $a_n$ and width $1$.

The diagram suggests that the partial sums of $\sum a_n$ satisfy the following inequality:

- $\displaystyle \sum_{n \mathop = 2}^N a_n \ge \frac 1 2 \sum_{n \mathop = 2}^N 2^n a\left({2^n}\right)$

- $\iff \displaystyle \sum_{n \mathop = 2}^N 2a_n \ge \sum_{n \mathop = 2}^N 2^n a\left({2^n}\right)$

To formalize this claim, observe that because $\langle {a_n}\rangle$ is decreasing:

\(\displaystyle \sum_{n \mathop = 2}^N 2a_n\) | \(=\) | \(\displaystyle a_2 + \underbrace{a_2 + a_3}_{\ge 2a_4} + \underbrace{a_3 + 2a_4 + a_5}_{\ge 4a_8} + \underbrace{a_5 + 2a_6 + 2a_7 + 2a_8 + a_9}_{\ge 8a_{16} } + \cdots + 2a_N\) | |||||||||||

\(\displaystyle \) | \(\ge\) | \(\displaystyle a_2 + 2a_4 + 4a_8 + 8a_{16} + \cdots + 2^n a \left({2^n}\right)\) |

By hypothesis, the left hand side converges as $n \to +\infty$.

Hence the condensed series converges as well, by the Comparison Test.

$\blacksquare$

## Source of Name

This entry was named for Augustin Louis Cauchy.

## Also see

## Sources

- 2009: Steven G. Krantz:
*Discrete Mathematics Demystified*: $\S 13.10$