Lecture 16 - The Series Tests

Exam II is coming up next week, so be mindful of those.

#todo/school : Add stuff to Cauchy Criterion once added in Lecture 15 - Cauchy Convergence 📅 2024-10-21 ✅ 2024-10-28

Limit Laws (Algebraic Limit Theorem) for Series

If $\sum a_{n} = A$ and $b = \sum b_{n} = B$ and $c \in R$ . then:

$\sum (c a_{n}) = c A$
$\sum (a_{n} + b_{n}) = A + B$

Proof

We must argue that the sequence of partial sums:
$t_{m} = c a_{1} + c a_{2} + \dots + c a_{m}$
converges to $c A$ . We are just given $\sum_{k = 1}^{\infty} a_{k}$ converges to $A$ . So the partial sums:
$s_{m} = a_{1} + a_{2} + \dots + a_{m}$
converge to $A$ . Because $t_{m} = c s_{m}$ then using the Limit Laws (Algebraic Limit Theorem) yields $(t_{m}) \to c A$
It's proved in very much the same way, using the summation ALT.
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We also talked about Cauchy Criterion for Series:

Cauchy Criterion for Series

$\sum a_{n}$ converges iff $\forall ε > 0 \exists N \in N$ such that for all $n > m \geq N$ then:
$| \sum_{k = m + 1}^{n} a_{k} | = \underset{s_{n} - s_{m}}{\underset{⏟}{| a_{m + 1} + a_{m + 2} + \dots + a_{n} |}} < ε$

Proof

Observe that:
$| s_{n} - s_{m} | = | a_{m + 1} + \dots + a_{n} |$
and apply the Cauchy Criterion.

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Notice there's no condition on the $a_{n}$ 's.

Example: Test for Divergence

Test for Divergence

If $\sum a_{n}$ converges then $a_{n} \to 0$ .

Proof
Let $ε > 0$ . By the Cauchy Condition for series, then choose $N \in N$ such that $n > m \geq N$ implies:
$| a_{m + 1} + \dots + a_{n} | < ε$
In particular, use $m = n - 1$ and any $n > N$ then:
$| a_{n} | = | a_{n} - 0 | < ε$
So by definition then $a_{n} \to 0$ .
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Example: Comparison Test

Comparison Test

Suppose $\forall n \in N$ we have $a_{n}, b_{n} \in R$ satisfy $0 \leq a_{n} \leq b_{n}$ . Then:

If $\sum b_{n}$ converges, then $\sum a_{n}$ converges.
If $\sum a_{n}$ diverges, then $\sum b_{n}$ diverges (automatically from (1))

Proof
Suppose $\sum b_{n}$ converges. Let's show $\sum a_{n}$ satisfies the Cauchy Condition for Series, and thus converges.

Let $ε > 0$ . Since $\sum b_{n}$ converges then we can choose $N \in N$ such that $n > m \geq N$ has it that:
$| b_{m + 1} + \dots + b_{n} | < ε$
Everything here is non-negative, so we can get rid of the absolute value signs:
$b_{m + 1} + \dots + b_{n} < ε$
Suppose this is true, that we choose our $n > m \geq N$ . Then:
$\begin{aligned} | a_{m + 1} + \dots + a_{n} | & = a_{m + 1} + \dots + a_{n} \\ \leq b_{m + 1} + \dots + b_{n} \\ < ε \end{aligned}$
Thus $\sum a_{n}$ satisfies the Cauchy Condition for Series, so then it must converge.
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Example: Absolute Convergence Test

ACT (Absolute Convergence Test)

If $\sum | a_{n} |$ converges, then $\sum a_{n}$ converges.

Proof
Let $ε > 0$ . Since $\sum | a_{n} |$ converges, then we can choose $N \in N$ such that any $n > m \geq N$ implies:
$| | a_{m + 1} | + \dots + | a_{n} | | < ε$
Choose any $n > m \geq N$ . Then:
$\begin{aligned} | a_{m + 1} + \dots + a_{n} | & \leq | a_{m + 1} | + \dots + | a_{n} | & Δ Inequality \\ = | | a_{m + 1} | + \dots + | a_{n} | | \\ < ε \end{aligned}$
Thus the $\sum a_{n}$ is Cauchy, so then $\sum a_{n}$ converges.
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Types of Convergence

Let $\sum a_{n}$ :

If $\sum | a_{n} |$ converges, then $\sum a_{n}$ is absolutely convergent.
If $\sum a_{n}$ converges but $\sum | a_{n} |$ diverges, then $\sum a_{n}$ is conditionally convergent.
Otherwise $\sum a_{n}$ diverges.

For example:
$\sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{n}$
converges conditionally while:
$\sum_{n = 1}^{\infty} \frac{(- 1)^{n + 1}}{2^{n}}$ converges absolutely.

Alternating Series Test (AST)

Suppose $(a_{n})$ satisfies:

$(a_{n})$ is decreasing.
$a_{n} \to 0$
Then $\sum_{n = 1}^{\infty} (- 1)^{n + 1} (a_{n}) = a_{1} - a_{2} + a_{3} - a_{4} + \dots$ converges.

Proof
Both conditions gives that $a_{n} \geq 0$ (just use MCT). As a result, the proof is done in:

2.7.1

Question

Prove the Alternating Series Test (AST). This is done by showing that the sequence of partial sums:
$
s_{n} = a_{1} - a_{2} + a_{3} - \dots \pm a_{n}
$
converges. Different characterizations of completeness lead to different proof.
a. Prove the AST by showing that $(s_{n})$ is Cauchy.
b. Supply another proof for this result using the Nested Interval Property.
c. Consider the subsequences $(s_{2 n}), (s_{2 n + 1})$ and show how the MCT leads to a third proof for the AST.

Scratch Work:

a.
For scratch, if we have this case then:
$
\begin{align}
|s_{n} - s_{m}| &= |a_{1} - a_{2} + \dots \pm a_{n} - (a_{1} - a_{2} + \dots \pm a_{n} \mp \dots \pm a_{m})| \
&= \left|\sum_{k=n+1}^{m} (-1)^{n+1}a_{k}\right| \
& \leq \sum_{k=n+1}^{m} |a_{k}| & \text{ Triangle Ineqality} \
& < \sum_{k=n+1}^{m} \frac{\epsilon}{m-(n+1)} \
& = \epsilon
\end

Proof

a. Let $ϵ > 0$ . Now we know that $(a_{n}) \to 0$ so then since we can choose such an $N$ where:
$| a_{n} | < ϵ$
for some $n \geq N$ . Now let $m > n \geq N$ as follows. Then notice that $m - n > 0$ so then notice that each $| a_{n + 1} |, | a_{n + 2} |, \dots, | a_{n + (m - n)} | < \frac{ϵ}{m - n}$ since this chosen epsilon is greater than 0 (applying the limit definition above). Then:
$\begin{aligned} | a_{n + 1} - a_{n + 2} + \dots \pm a_{n + (m - n)} | & \leq | a_{n + 1} | + | - a_{n + 2} | + \dots + | \pm a_{n + (m - n)} \\ = | a_{n + 1} | + | a_{n + 2} | + \dots + | a_{n + (m - n)} | & (a_{n} \geq 0) \\ < (m - n) \cdot \frac{ϵ}{m - n} \\ = ϵ \end{aligned}$
But the LHS is just $| s_{n} - s_{m} |$ so we've really shown:
$| s_{n} - s_{m} | < ϵ$
showing it's Cauchy. Thus $(s_{n})$ converges, showing the AST.

b. Consider constructing the intervals $I_{n}$ such that for each $n$ :
$I_{n} = {\begin{cases} [s_{n}, s_{n + 1}] & n is even \\ [s_{n + 1}, s_{n}] & n is odd \end{cases}$
The alternation is well defined since when $n = 2 k$ ( $n$ is even):
$s_{n + 1} - s_{n} = (- 1)^{2 k + 1 + 1} a_{n + 1} = a_{n + 1} \geq 0$
and for when $n = 2 k + 1$ ( $n$ is odd) then you get $s_{n} \geq s_{n + 1}$ in a similar way. This shows these intervals are always defined. Further, notice we can show that $I_{n + 1} \supset I_{n}$ . Notice first that since $a_{n}$ is decreasing (and positive), then $a_{n + 1} \leq a_{n}$ . That means that:
$a_{n + 1} - a_{n} \leq 0 ⟺ s_{n + 1} - s_{n - 1} \leq 0 ⟺ s_{n + 1} \leq s_{n - 1}$
so the series of partial sums are also decreasing as a result (in two term intervals):

s_{n + 1} \leq s_{n - 1}

for all

n

Let $x \in I_{n}$ . We'll show that $x \in I_{n + 1}$ :

If $n$ is even then $x \in [s_{n}, s_{n + 1}]$ and then $n + 1$ is odd so we'll show $x \in [s_{n + 2}, s_{n + 1}]$ . Notice that already $x \leq s_{n + 1}$ . For the other inequality: $x \geq s_{n} \geq s_{n + 2}$ per our above lemma.
If $n$ is odd, the proof comes out very similarly.

Thus we've shown the required nested interval property, so then:
$\exists x s.t. x \in ⋂_{n = 1}^{\infty} I_{n}$
Now we really want to show that $(s_{n}) \to x$ . Let $ϵ > 0$ . Since $(a_{n}) \to 0$ then we know that $\exists N \in N$ where any $n \geq N$ has it that $| a_{n} | < ϵ$ . Now notice that if we let any $n \geq N$ :
$\begin{aligned} | s_{n} - x | & \leq | s_{n} - s_{n + 1} | & x is at most/least from the interval’s endp. \\ = | - a_{n + 1} | \\ = | a_{n + 1} | \\ < ϵ \end{aligned}$
Thus completing the proof.

c. Notice that the subsequence $(s_{2 n}), (s_{2 n + 1})$ all are decreasing (see the lemma from (b)) and bounded below (given by how $a_{n} \geq 0$ ). As a result, then by the MCT both subsequences converge. Now by the Limit Laws (Algebraic Limit Theorem) then because:
$(s_{n}) \equiv (s_{2 n} + s_{2 n + 1})$
Then $(s_{n})$ converges to the same limit as the sum of the limits of the two subsequences.

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Example: The Alternating Harmonic Series

Recall the AHS:

1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots

this series converges by the AST; however, one of our cautionary tales said that we could reorder the terms to get a new number. Wouldn't that contradict it being convergent (since then the limits wouldn't be unique).

Rearrangement

Given $\sum_{n = 1}^{\infty} a_{n}$ and a bijection $f : N \to N$ . Define $b_{f (k)} = a_{k}$ . Then $\sum_{n = 1}^{\infty} b_{n}$ is a rearrangement of $\sum_{n = 1}^{\infty} a_{n}$ .

Rearrangement Theorem

If $\sum a_{n}$ is conditionally convergent, and $α \in R$ . Then $\exists$ a rearrangement $\sum b_{k}$ such that $\sum_{k = 1}^{\infty} b_{k} = α$ .

Proof
Think of a proof outline for the AST. If you only add the positive terms, it must diverge, and similarly for the negative terms. Similarly, in a conditionally convergent series you can use the positive terms, add one negative term, then add positive terms until you get too far, then add one negative term, and rinse and repeat:

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Theorem

If $\sum a_{n}$ is absolutely convergent, then any rearrangement converges to the same value.

Proof
Let $f : N \to N$ where $b_{f (k)} = a_{k}$ so that $\sum b_{k}$ is a rearrangment. Then let:
$s_{n} = a_{1} + \dots + a_{n}$
$t_{m} = b_{1} + \dots + b_{m}$
Since $\sum a_{n}$ is absolutely convergent, then if we let $ε > 0$ to show convergence then:
$\sum a_{n} = A$
By our supposition. We can choose $N \in N$ such that for any $n \geq N$ that:
$| s_{n} - A | < \frac{ε}{2}$
Similarly by the Cauchy Criterion, then $\forall n > m \geq N$ then:
$| a_{m + 1} | + \dots + | a_{n} | < \frac{ε}{2}$
Let $M = max (f (1), f (2), \dots, f (N))$ . If we choose $m \geq M$ then $m \geq f (i)$ so then:
$| t_{m} - A | \leq \underset{\sum a_{k} : k > N}{\underset{⏟}{| t_{m} - s_{N} |}} + \underset{< \frac{ε}{2}}{\underset{⏟}{| s_{N} - A |}}$
But since the Cauchy Criterion makes the first term $< \frac{ε}{2}$ , finishing the proof.
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