27 Properties of Infinite Series Practice

#todo/school : Do problems from 2.7 # 1, 2, 3b, 4, 7, 8, 9 ✅ 2024-10-29

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{aligned} | 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}) | \\ = | \sum_{k = n + 1}^{m} (- 1)^{n + 1} a_{k} | \\ \leq \sum_{k = n + 1}^{m} | a_{k} | & Triangle Ineqality \\ < \sum_{k = n + 1}^{m} \frac{ϵ}{m - (n + 1)} \\ = ϵ \end{aligned}

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.

☐

2.7.2

Question

Decide whether each of the following series converges or diverges:
a. $\sum_{n = 1}^{\infty} \frac{1}{2^{n} + n}$
b. $\sum_{n = 1}^{\infty} \frac{\sin (n)}{n^{2}}$
c. $1 - \frac{3}{4} + \frac{4}{6} - \frac{5}{8} + \frac{6}{10} - \frac{7}{12} + \dots$
d. $1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9} + \dots$
e. $1 - \frac{1}{2^{2}} + \frac{1}{3} - \frac{1}{4^{2}} + \frac{1}{5} - \frac{1}{6^{2}} + \frac{1}{7} - \frac{1}{8^{2}} + \dots$

Proof

a. Comparison Test (Series) with $\sum \frac{1}{2^{n}}$ which is a convergent geometric series. Thus the series converges.

b. Comparison Test (Series) with $\sum \frac{1}{n^{2}}$ which is a convergent $p = 2$ series. Thus the series converges.

c. Equivalent to asking if $\sum_{n = 1}^{\infty} (- 1)^{n + 1} \frac{n + 1}{2 n}$ converges. Using Alternating Series Test (AST) notice that:

\begin{aligned} a_{n + 1} \leq a_{n} & ⟺ \frac{n + 2}{2 (n + 1)} \leq \frac{n + 1}{2 n} \\ ⟺ 2 n^{2} + 4 n \leq 2 n^{2} + 4 n + 2 \\ ⟺ 0 \leq 2 \end{aligned}

Is always true, showing the decreasing part of the test. Now to show that $\frac{n + 1}{2 n} \to 0$ but this is false, so this implies the series diverges (but we need to use a different test). Test for divergence will work here since if we consider the two subsequences $a_{2 n}$ and $a_{2 n + 1}$ then it's clear that $a_{2 n} \to - \frac{1}{2}$ while $a_{2 n + 1} \to \frac{1}{2}$ so then since the subsequences have different limits then $(a_{n})$ doesn't work, thus the Test for Divergence (Series) shows the series diverges.

d. This is (almost) the alternating harmonic series. This is equivalent to asking if the series:

(1 + \frac{1}{2}) - \frac{1}{3} + (\frac{1}{4} + \frac{1}{5}) - \frac{1}{6} + \dots \equiv \sum_{n = 1}^{\infty} ((\frac{1}{3 n - 2} + \frac{1}{3 n - 1}) - \frac{1}{3 n})

converges. The sequence used in the series reduces down to:

\frac{3 n - 1 + 3 n - 2}{(3 n - 2) (3 n - 1)} - \frac{1}{3 n} = \frac{9 n^{2} - 2}{3 n (3 n - 1) (3 n - 2)}

We could direct comparison with a standard $p = 1$ (harmonic series) and get that the series diverges.

e. Consider the two subsequence used for the series $a_{2 n - 1} = \frac{1}{2 n - 1}$ and $a_{2 n} = - \frac{1}{(2 n)^{2}}$ .
Assume for contradiction that this converges. Then:

\sum_{n = 1}^{\infty} a_{n} = \sum_{n = 1}^{\infty} a_{2 n} - \sum_{n = 1}^{\infty} a_{2 n - 1} = \sum_{n = 1}^{\infty} \frac{1}{2 n - 1} - \sum_{n = 1}^{\infty} \frac{1}{(2 n)^{2}}

But look! The first series diverges by a comparison test with the harmonic series, while we assumed it converged. Thus, our original series must have diverged.

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2.7.3

Question

a. Provide the proof of the Comparison Test (Series) using the Cauchy Criterion for Series.
b. Give another proof for the Comparison Test, this time using the MCT.

Proof

a. See Comparison Test (Series) as the proof is already provided.

b. Proving (1) is the only real thing for this proof. Suppose $\sum b_{n}$ converges and $0 \leq a_{n} \leq b_{n}$ for some unknown arbitrary $a_{n}$ sequence. We'll show that the series of partial sums of $\sum a_{n}$ converges, namely that the sequence:

s_{m} = a_{1} + a_{2} + \dots + a_{m}

converges. Note that since $0 \leq a_{n}$ then $s_{m}$ is strictly increasing, so it's monotone. We only need to show that $s_{m}$ is bounded above. Notice that $\sum b_{n} \to L$ for some $L$ so since $s_{m} \leq t_{m}$ where $t_{m}$ is the convergent partial sum of $b_{n}$ , then by the Limits and Order (Order Limit Theorem) then $s_{m} \leq L$ as an upper bound.

Thus, the MCT says that $s_{m}$ must converge.
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2.7.4

Question

Give an example or explain why the request is impossible referencing the proper theorems.
a. Two series $\sum x_{n}$ and $\sum y_{n}$ both diverge but $\sum x_{n} y_{n}$ converges.
b. A convergent series $\sum x_{n}$ and a bounded sequence $(y_{n})$ such that $\sum x_{n} y_{n}$ diverges.
c. Two sequences $(x_{n}), (y_{n})$ where $\sum x_{n}$ and $\sum (x_{n} + y_{n})$ both converge but $\sum y_{n}$ diverges.
d. A sequence $(x_{n})$ where $0 \leq x_{n} \leq \frac{1}{n}$ where $\sum (- 1)^{n} x_{n}$ diverges.

Proof

a. Possible. Let $x_{n} = y_{n} = \frac{1}{n}$ . Clearly $\sum \frac{1}{n}$ diverges but $\sum \frac{1}{n^{2}}$ converges.

b. Let $x_{n} = \frac{(- 1)^{n}}{n}$ and $y_{n} = (- 1)^{n}$ . Then $\sum \frac{(- 1)^{n}}{n} (- 1)^{n} = \sum \frac{1}{n}$ diverges.

c. Impossible. If $\sum x_{n}$ and $\sum (x_{n} + y_{n})$ converge then $\sum [(x_{n} + y_{n}) - x_{n}] = \sum y_{n}$ must converge.

d. If we have:

x_{n} = {\begin{cases} \frac{1}{n} & n is even \\ 0 & n is odd \end{cases}

Then essentially this diverges in the same way the harmonic series does.
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2.7.7

Question

a. Show that if $a_{n} > 0$ and $lim_{n \to \infty} (n a_{n}) \to l$ with $l \neq 0$ then the series $\sum a_{n}$ diverges.
b. Assume $a_{n} > 0$ and $lim_{n \to \infty} (n^{2} a_{n})$ exists. Show $\sum a_{n}$ converges.

Scratch:

a. Notice that $a_{n} \sim \frac{l}{n} = l \cdot \frac{1}{n}$ . This implies that we should show $a_{n} \geq c \cdot \frac{1}{n}$ . If we go far enough out and choose $N$ where any $n \geq N$ has:

| n a_{n} - l | < ϵ

but we know $\frac{l}{2} > 0$ , so then:

| n a_{n} - l | < \frac{l}{2} \Rightarrow \frac{- l}{2} > n a_{n} - l > \frac{l}{2} \Rightarrow n a_{n} > \frac{l}{2} \Rightarrow a_{n} > \frac{l}{2} \cdot \frac{1}{n}

so $\sum_{k = N}^{\infty} a_{k}$ diverges by the Comparison Test (Series) (with a Harmonic Series).

b. It's a similar idea except you use the higher inequality.

Proof

a. Since $(n a_{n}) \to l$ then for any $ϵ > 0$ then we can choose some $N \in N$ such that any $n \geq N$ gives:

| n a_{n} - l | < ϵ

now notice $\frac{| l |}{2} > 0$ , so then:

\begin{array}{r} | n a_{n} - l | < \frac{| l |}{2} \\ \frac{- | l |}{2} < n a_{n} - l < \frac{| l |}{2} \\ l - \frac{| l |}{2} < n a_{n} < l + \frac{| l |}{2} \\ \frac{l - \frac{| l |}{2}}{n} < a_{n} < \frac{l + \frac{| l |}{2}}{n} \end{array}

Thus direct comparison test with $\sum \frac{l - \frac{| l |}{2}}{n}$ shows that since this sum diverges (harmonic series) then $\sum a_{n}$ diverges.

b. Since $(n^{2} a_{n}) \to l$ then we have:

| n^{2} a_{n} - l | < ϵ

for any $n \geq N$ for some choosable $N$ . Now notice that $| l | > 0$ so using that as our epsilon:

\begin{array}{r} | n^{2} a_{n} - l | < | l | \\ - | l | < n^{2} a_{n} - l < | l | \\ l - | l | < n^{2} a_{n} < l + | l | \\ \frac{l - | l |}{n^{2}} < a_{n} < \frac{l + | l |}{n^{2}} \end{array}

So direct comparison with $\sum \frac{l + | l |}{n^{2}}$ which is a convergent $p = 2$ series shows that $\sum a_{n}$ converges.

☐

2.7.8

Question

Consider each of the following propositions. Provide short proofs for those that are true and counterexamples for those that are not.
a. If $\sum a_{n}$ converges absolutely, then $\sum a_{n}^{2}$ converges absolutely.
b. If $\sum a_{n}$ converges and $(b_{n})$ converges, then $\sum a_{n} b_{n}$ converges.
c. If $\sum a_{n}$ converges conditionally, then $\sum n^{2} a_{n}$ diverges.

Proof

a. True. If $\sum a_{n}$ converges absolutely then $a_{n} \to 0$ so eventually $a_{n}^{2} \leq | a_{n} |$ so then comparing this with the convergent $\sum | a_{n} |$ means that $\sum a_{n}^{2}$ must converge. It has to absolutely since all the sequence terms are positive.

b. False. Use both $a_{n} = b_{n} = \frac{(- 1)^{n}}{\sqrt{n}}$ . The goal is to use conditional convergence here.

c. True. Here then $\sum | a_{n} |$ diverges. The goal is to compare $n^{2} a_{n}$ with $| a_{n} |$ and show it's bigger:

n^{2} a_{n} = | n a_{n} |^{2} \geq | n a_{n} | \geq | a_{n} |

☐

2.7.9

Ratio Test

Given a series $\sum a_{n}$ with $a_{n} \neq 0$ the Ratio Test states that if $(a_{n})$ satisfies:

lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = r < 1

then the series converges absolutely.
a. Let $r^{'}$ satisfy $r < r^{'} < 1$ . Explain why $\exists N$ such that $n \geq N$ implies $| a_{n + 1} | \leq | a_{n} | r^{'}$ .
b. Why does $| a_{N} | \sum (r^{'})^{n}$ converge?
c. Now show that $\sum | a_{n} |$ converges, and conclude that $\sum a_{n}$ converges.

Scratch:

a. For (a) we want:

\begin{aligned} | a_{n + 1} | & \leq | a_{n} | r^{'} \\ | \frac{a_{n + 1}}{a_{n}} | & \leq r^{'} \\ | \frac{a_{n + 1}}{a_{n}} | - r & \leq r^{'} - r \\ | | \frac{a_{n + 1}}{a_{n}} | - r | & \leq | r^{'} - r | \end{aligned}

Hey we should use our $ϵ = | r^{'} - r | > 0$ . We can lose the absolute values though since $r^{'} - r > 0$ by our supposition.

Proof

a. The limit says that for any $ϵ > 0$ we can choose such an $N$ where any $n \geq N$ implies:

| | \frac{a_{n + 1}}{a_{n}} | - r | < ϵ

Now since $r^{'} - r > 0$ by our supposition then it goes to see that:

\begin{aligned} | | \frac{a_{n + 1}}{a_{n}} | - r | & < r^{'} - r \\ - (r^{'} - r) & < | \frac{a_{n + 1}}{a_{n}} | - r < r^{'} - r \end{aligned}

Using only the right inequality:

\Rightarrow | \frac{a_{n + 1}}{a_{n}} | < r^{'} \Rightarrow | a_{n + 1} | < r^{'} | a_{n} |

as desired. The equals case only happens our ratio of terms is some constant, which could happen.

b. Since $r^{'} < 1$ then the series $\sum (r^{'})^{n}$ is a geometric series that has to converge. As a result since we can use the same $N$ , then $| a_{N} |$ is some constant, so $| a_{N} | \sum (r^{'})^{n}$ must converge.

c. Since $n \geq N$ , notice that from repeated use of (a):

| a_{n} | < r^{'} | a_{n - 1} | < r^{' 2} | a_{n - 2} | < \dots < (r^{'})^{n - N} | a_{N} |

This implies that:

\begin{aligned} | a_{N} | + | a_{N + 1} | + \dots + | a_{n} | & < | a_{N} | + r^{'} | a_{N} | + \dots + (r^{'})^{n - N} | a_{N} | \\ \sum_{k = N}^{n} | a_{k} | & < | a_{N} | \sum_{k = 0}^{n - N} (r^{'})^{k} \end{aligned}

Looking at our original series we want to see $\sum | a_{n} |$ converges, looking at the partial sums:

s_{n} = \sum_{k = 1}^{n} | a_{k} | = \sum_{k = 1}^{N - 1} | a_{k} | + \sum_{k = N}^{n} | a_{k} | < \sum_{k = 1}^{N - 1} | a_{k} | + | a_{N} | \sum_{k = 0}^{n - N} (r^{'})^{k}

Now the rightmost series converges since it's a geometric series and $| a_{N} |$ is constant. The series before it is finite in number of terms ignorant of $n$ (and thus convergent), so then $s_{n}$ must converge if we take the limits of both sides and do the Limits and Order (Order Limit Theorem).

☐