Cauchy Condensation Test

Theorem
Let $\left \langle {a_n} \right \rangle: n \mapsto f\left({n}\right)$ 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 $\left \langle {a_n} \right \rangle$: $a_n > 0, a_{n+1} \le a_n$, and $a_n \to 0$ as $n \to +\infty$.

Then the series:


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

converges iff the series:


 * $\displaystyle \sum_{n=1}^\infty 2^n f\left({2^n}\right)$

converges.

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

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

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


 * Cauchycondensation1.png

The dotted black line represents the sequence of partial sums of $\sum a_n$.

The $n$th rectangle has:


 * Base equal to $2^n$


 * Altitude equal to $f\left({2^n}\right)$

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


 * $\displaystyle \sum_1^N 2^n f\left({2^n}\right)$

which is precisely the defined condensed series.

It is plain from the picture that the partial sums of $\sum a_n$ are not greater than the condensed partial sums:


 * $\displaystyle \sum_{n=1}^N a_n \le \sum_{n=1}^\infty 2^n f\left({2^n}\right)$

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

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