Cauchy Condensation Test

From Series, Convergence of A Series#Example 4 The General Case we have:

Example 4: The General Case

Now consider the abstract, general case of $\forall n \in N$ where $a_{n} \geq 0$ and $a_{n + 1} \leq a_{n}$ . Let $s_{m} = \sum_{n = 1}^{m} a_{n}$ . Then let $m = 2^{k}$ similar to our previous example:
$\begin{aligned} s_{2^{k}} & = a_{1} + a_{2} + (a_{3} + a_{4}) + (a_{5} + \dots + a_{8}) + \dots + (a_{2^{k - 1} + 1} + \dots + a_{2^{k}}) \\ \geq a_{1} + a_{2} + 2 a_{4} + 4 a_{8} + \dots + 2^{k - 1} a_{2^{k}} \\ \geq \frac{1}{2} \underset{:= t_{k}}{\underset{⏟}{(a_{1} + 2 a_{2} + 4 a_{4} + \dots + 2^{k} a_{2^{k}})}} \end{aligned}$
So if $m \geq 2^{k}$ then $s_{m} \geq \frac{1}{2} t_{k}$ . Showing that $t_{k}$ is unbounded would imply that $s^{2^{k}}$ is unbounded, showing a divergent series.

We could do a bound in the other direction too:
$\begin{aligned} s_{2^{k}} & = a_{1} + \underset{\leq 2 a_{2}}{\underset{⏟}{(a_{2} + a_{3})}} + \underset{\leq 4 a_{4}}{\underset{⏟}{(a_{4} + \dots + a_{7})}} + \underset{\leq 8 a_{8}}{\underset{⏟}{(a_{8} + \dots + a_{15})}} + \dots + \underset{\leq 2^{k - 1} a_{2^{k - 1}}}{\underset{⏟}{(\frac{1}{a_{2^{k - 1}}} + \dots + \frac{1}{a_{2^{k} - 1}}) + a^{2^{k}}}} \\ \leq a_{1} + 2 a_{2} + 4 a_{4} + 8 a_{8} + \dots + 2^{k - 1} a_{2^{k - 1}} + 2^{k} a_{k} \\ = t_{k} \end{aligned}$
So if $m \leq 2^{k}$ then $s_{m} \leq t_{k}$ .

This gives the following:

Cauchy Condensation Test

Suppose $\forall n \in N$ that $a_{n} \geq 0$ and $a_{n} \geq a_{n + 1}$ . Then:

\sum_{n = 1}^{\infty} a_{n} converges \equiv \sum_{m = 0}^{\infty} 2^{m} a_{2^{m}} converges

Proof
Let $a_{n}$ , $n \in N$ be as in the theorem. For $ℓ, k \in N$ set:

s_{ℓ} = a_{1} + a_{2} + \dots + a_{ℓ}

t_{k} = a_{1} + 2 a_{2} + 4 a_{4} + \dots + 2^{k} a_{2^{k}}

For the $\to$ suppose $\sum_{n = 1}^{\infty} a_{n}$ converge.s. Then $\exists s \in R$ such that $s_{ℓ} \leq s$ for all $ℓ \in N$ . Then $\forall k \in N$ we have (using $\frac{1}{2} t_{k} \leq s_{2^{k}} \leq t_{k}$ and multiplying all sides by 2):

t_{k} \leq 2 s_{2^{k}} \leq 2 s

Thus $t_{k}$ is bounded, so then $\sum_{m = 0}^{\infty} 2^{m} a_{2^{m}}$ 's partial sums are bounded, so it must converge.
2. For the $\leftarrow$ suppose $\sum_{m = 0}^{\infty} 2^{m} a_{2^{m}}$ converges. Then it has some bound $t \in R$ where $\forall k \in N$ we have $t_{k} \leq t$ . Let $ℓ \in N$ . We can always pick some $k \in N$ such that $ℓ \leq 2^{k}$ . Then:

s_{ℓ} \leq s_{2^{k}} \leq t_{k} \leq t

So then $s_{ℓ}$ is bounded, implying that $\sum_{n = 1}^{\infty} a_{n}$ is bounded and thus convergent.
☐

Example: $p$ -Series

p

-Series

The series $\sum_{n = 1}^{\infty} \frac{1}{n^{p}}$ converges iff $p > 1$ .

Let $p \in R$ . Determine the convergence for:

\sum_{n = 1}^{\infty} \frac{1}{n^{p}}

If $p \leq 0$ then $\frac{1}{n^{p}} \geq 1$ so then the partial sums are $\geq \underset{n times}{\underset{⏟}{1 + 1 + \dots + 1}} = n$ which is a divergence sequence. Hence the series diverges in this case.

Now suppose instead $p > 0$ . Then CCT applies:

\sum_{n = 1}^{\infty} \frac{1}{n^{p}} converges \equiv \sum_{n = 0}^{\infty} 2^{m} {(\frac{1}{2^{m}})}^{p} converges

But:

2^{m} {(\frac{1}{2^{m}})}^{p} = \frac{2^{m}}{2^{m p}} = \frac{1}{2^{m p - m}} = {(\frac{1}{2^{p - 1}})}^{m}

So the series becomes:

= \sum_{n = 0}^{\infty} {(\frac{1}{2^{p - 1}})}^{m}

Is a Series, Convergence of A Series#Example Geometric Series, so then this converges iff $\frac{1}{2^{p - 1}} < 1 \equiv p > 1$ .

Thus, our $p$ -series converges iff $p > 1$ .

Example 4: The General Case

Example: p-Series

Example: $p$ -Series