Lecture 14 - Weierstrass Convergence Theorem

Last time we talked about Series, Convergence of A Series. In that we talked about how:

\frac{1}{2} t_{k} \leq s_{2^{k}} \leq t_{k}

This is the basis of showing the 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\limits_{m=0}^{\infty} 2^{m}a_

Then we talked about Subsequence Convergence Theorem:

subsequence

Let $(a_{n})$ be a sequence. Suppose that:
$n_{1}, n_{2}, n_{3}, \dots \in N$
satisfying:
$1 \leq n_{1} < n_{2} < n_{3} < \dots$
Then the sequence $(a_{n_{k}})$ for all $k \in N$ is a subsequence of $(a_{n})$ . It's a proper subsequence if $(a_{n_{k}})$ doesn't have all $k = n_{k}$ , namely $\exists k$ where $k \neq n_{k}$ .

The idea is that we just remove possibly some terms, and then preserve the order.

\forall k \in N, n_{k} \geq k

Example

Consider the sequence:
$(\frac{1}{n}) = (1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots)$
Some sub sequences of these include:
$(1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \dots)$
$(1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \dots)$
$(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \dots)$
an invalid sub sequence would be:
$(1, \frac{1}{3}, \frac{1}{2}, \frac{1}{4}, \dots)$
since the order of terms in the sequence must be preserved.

Notice that each of these sub sequences converges to the same value (namely 0 in this case) to the original sequence. This is what the next theorem tries to show.

The Theorem

Subsequence Convergence Theorem

A subsequence of a convergent sequence converges to the same value as the original sequence.

Proof
Note our observation that $\forall k \in N, n_{k} \geq k$ . Say $(a_{n})$ converges to some $L$ . Then $\forall ε > 0 \exists N \in N$ where for all $n \geq N$ then:
$| a_{n} - L | < ε$
Let $(a_{n_{k}})$ be a subsequence. Let $ε > 0$ . Then $\exists N \in N$ where for any $n \geq N$ then $| a_{n} - a | < ε$ . Suppose $k \geq N$ . Then by construction then $n_{k} \geq k \geq N$ so then $| a_{n_{k}} - a | < ε$ as desired.
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Example of WCT

Consider the sequence:
$(1, - 1, 1, - 1, \dots)$
we could use the Convergence of a Sequence definition to show this diverges. But we can also show that the subsequences $a_{2 n}$ and $a_{2 n - 1}$ for all $n \in N$ are valid subsequences. They have different limits of -1 and 1 respectively. If $a_{n}$ itself converged, then these subsequences need to converge to the same limit since Limits are Unique. But they don't, so then by our SCT then the original sequence cannot converge.

See also: Balzano-Weierstrass Theorem.

Balzano-Weierstrass Theorem

Every bounded sequence has a convergent subsequence.

Scratch work:

Here we split each interval, and choose the one that has infinitely-many terms. Here we have it that:
$I_{1} \supseteq I_{2} \supseteq I_{3} \supseteq \dots$
We can pick an element $b_{1} \in I_{1}$ , then the next element $b_{2} \in I_{2}$ , and so on.

More formally, pick an $n_{1} \in N$ such that $a_{n_{1}} \in I_{1}$ , $n_{2} \in N$ where $a_{n_{2}} \in I_{2}$ and $n_{2} > n_{1}$ . Continue with $n_{3} \in N$ such that $a_{n_{3}} \in I_{3}$ and $n_{3} > n_{2}$ . Repeat as needed. We get that, by the Nested Interval Property, that
Proof

Let $(a_{n})$ be a bounded sequence so that $\exists M > 0$ where $| a_{n} | \leq M$ for all $n \in N$ . Bisect the closed interval $[- M, M]$ into the two closed intervals $[- M, 0]$ and $[0, M]$ (the midpoint is in both halves). Now it must be that at least one of these closed intervals contains an infinite number of the terms in the sequence $(a_{n})$ . Select a half for which this is the case and label that interval as $I_{1}$ . Then let $a_{n_{1}}$ be some term in the sequence $(a_{n})$ satisfying $a_{n_{1}} \in I_{1}$ .

Next we bisect $I_{1}$ into closed intervals of equal length and let $I_{2}$ be again the half that contains infinitely many terms from $(a_{n})$ . Select an $a_{n_{2}}$ from the original sequence with $n_{2} > n_{1}$ and $a_{n_{2}} \in I_{2}$ .

In general if we construct the closed interval $I_{k}$ by taking a half of $I_{k - 1}$ containing an infinite number of terms of $(a_{n})$ and then select $n_{k} > n_{k - 1} > \dots > n_{2} > n_{1}$ so that $a_{n_{k}} \in I_{k}$ .

We want to argue that $(a_{n_{k}})$ is a convergent subsequence. But we need a candidate for the limit. Via the Nested Interval Property since:
$I_{1} \supseteq I_{2} \supseteq \dots$
Then $\exists x \in R$ where $x \in I_{k}$ for every $k$ . It just remains to show that $(a_{n_{k}}) \to x$ . Let $ε > 0$ . Via 25 Subsequences and the Bolzano-Weierstrass Theorem Practice#2.5.3 and the Limit Laws (Algebraic Limit Theorem) then
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s partial sums are bounded, so it must converge. 2. For the $\leftarrow$ suppose $\sum\limits_{m=0}^{\infty} 2^{m}a_{2^{m}}$ converges. Then it has some bound $t \in \mathbb{R}$ where $\forall k \in \mathbb{N}$ we have $t_{k} \leq t$. Let $\ell \in \mathbb{N}$. We can always pick some $k \in \mathbb{N}$ such that $\ell \leq 2^{k}$. Then: $ s_{\ell} \leq s_{2^{k}} \leq t_{k} \leq t $ So then $s_{\ell}$ is bounded, implying that $\sum\limits_{n=1}^{\infty} a_{n}$ 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 $p > 1$ .

Then we talked about Subsequence Convergence Theorem:

!Subsequence Convergence Theorem

!Balzano-Weierstrass Theorem