Title: Math and science / Analysis / Tao / 07. Series
Tags: more_details_needed

Title: Finite series, definition
Q: Let \( m, n \) be integers, and let \( (a_i)_{i=n}^{m} \) be a finite sequence of real numbers, assigning a real number \( a_i \) for each integer \( i \) between \( n \) and \( m \) inclusive (i.e. \( m \le i \le n \)). Then we define the finite sum (or finite series) \( \sum_{i=m}^{n} a_i \) by the recursive formula
\[
\begin{aligned}
\sum_{i=m}^{n} a_i &= 0 \text{ whenever } n < m; \\
\sum_{i=m}^{n+1} a_i &= [...] \text{ whenever } [...].
\end{aligned}
\]
A: \left(\sum_{i=m}^{n} a_i\right) + a_{n+1}; n + 1 \ge m
BackOnlyText: The series is often less formally expressed as:
\( \sum_{n}^{i=m} a_i = a_m + a_{m+1} + ... + a_n \)
Semantically, "series" refers to the expression of the form \( \sum_{n}^{i=m} a_i \), which is mathematically, but not semantically equal to a real number, which is called the "sum" of the series. This is a linguistic distinction, and a distinction that is not pertinant in mathematics due to the axiom of substitution.
Example: \[
\begin{aligned}
\sum_{i=m}^{m-2} a_i &= 0\\
\sum_{i=m}^{m-1} a_i &= 0\\
\sum_{i=m}^{m} a_i &= a_m\\
\sum_{i=m}^{m+1} a_i &= a_m + a_{m+1} \\
\sum_{i=m}^{m+2} a_i &= a_m + a_{m+1} + a_{m+2}\end{aligned}\]
Source: p156-157

Title: Summation over finite sets, definition
Q: Let \( X \) be a set with \( n \in \mathbb{N} \) elements and let \( f: X \rightarrow \mathbb{N} \) be a function from \( X \) to the real numbers. Then we define the finite sum \( \sum_{x \in X} f(x) \) as follows.
Select any [...] \( g \) from \( \{i \in \mathbb{N} : 1 \le i \le n \} \) to \( X \). Such a [...] exists since \( X \) is assumed to have \( n \) elements. We then define:
\[ \sum_{x \in X} f(x) := \sum_{i=1}^{n}[...] \]
A: bijection; bijection; f(g(i))
BackOnlyText: While this is just a definition, we still need to check that it creates a valid identity that is consistent with the requirements of equality. Specifically, the requirements of the substitution axiom must be met. Our definition allows for freedom to choose a bijection, so we must insure that any such choice of bijection produces a sum equal to a sum produced from another bijection. This is somewhat more involved that one might expect.
The proof of the following proposition is a good example of a non-obvious induction setup (induction over the size of a set is used). Refer to the book for the proof, as it is symbolically involved.
Proposition, finite summations are well-defined
Let \( X \) be a finite set with \( n \) elments (where \( n \in \mathbb{N} \)), let \( f: X \rightarrow \mathbb{R} \) be a function, and let \( g: \{i \in \mathbb{N} : 1 \le i \le n \} \rightarrow X \) and \( h: \{i \in \mathbb{N} : 1 \le i \le n \} \rightarrow X \) be bijections. Then we have
\[ \sum_{i=i}^{n}f(g(i)) = \sum_{i=2}^{n}f(h(i)) \]
Example: Let \( X \) be the set \( X := \{a, b, c\} \). Let \( f:X \rightarrow X \) be the function \( f(a) := 2 \), \( f(b) := 5 \) and \( f(c) := -1 \). Select a bijection \( g: \{1, 2, 3\} \rightarrow X \), for example \( g(1) := a \), \( g(2) := b \) and \( g(3) := c \). Then we have:
\[
\begin{aligned}
\sum_{x \in X}f(x) &= \sum_{i=1}^{3}f(g(i)) \\
&= f(a) + f(b) + f(c) \\
&= 6
\end{aligned} \]

Title: Series laws Ⅱ: sums over finite sets
9 basic properties of sums over finite sets. The proofs of these are a good exercise to refresh the notions of functions, bijections, sets, sums and induction.
There is 1-1 correspondence between many of these properties and the 6 properties already shown for sums of the form \( \sum_{i=1}^{n}a_i \). Thus, proving many of them amounts to choosing a bijection that allows transitioning to one of the original properties.
Question: when reading the below properties, consider:
- what is the equivalent \( \sum_{i=1}^{n}a_i \) form, if it exists?
- does the property extend to sums over infinite sets?
1. Empty set. If \( X \) is empty, and \( f: X \rightarrow \mathbb{R} \) is a function (i.e., \( f \) is the empty function), we have
\[ \sum_{x \in X} f(x) = 0 \]
2. Single element set. If \( X \) consists of a single element, \( X = \{x_0\} \), and \( f: X \rightarrow \mathbb{R} \) is a function, we have
\[ \sum_{x \in X}f(x) = f(x_0) \]
3. Substitution, part I. If \( X \) is a finite set, \( f: X \rightarrow \mathbb{R} \) is a function, and \( g: Y \rightarrow X \) is a bijection, then
\[ \sum_{x \in X}f(x) = \sum_{y \in Y}f(g(y)) \]
4. Substitution, part II. Let \( n \le m \) be integers, and let \( X \) be the set \( X := \{i \in \mathbb{Z} : n \le i \le m\} \). If \( a_i \) is a real number assigned to each integer \( i \in X \), then we have
\[ \sum_{i=n}^{m}a_i = \sum_{i \in X}a_i \]
5. Disjoint sets. Let \( X \), \( Y \) be disjoint finite sets (\( X \cap Y = \emptyset \)), and \( f: X \cup Y \rightarrow \mathbb{R} \) is a function. Then we have
\[ \sum_{z \in X \cup Y}f(z) = \left( \sum_{x \in X} f(x) \right) + \left( \sum_{y \in Y} f(y) \right) \]
6. Linearity, part I. Let \( X \) be a finite set, and let \( f: X \rightarrow \mathbb{R} \) and \( g: X \rightarrow \mathbb{R} \) be functions. Then we have
\[ \sum_{x \in X}(f(x) + g(x)) = \sum_{x \in X}f(x) + \sum_{x \in X}g(x) \]
7. Linearity, part II. Let \( X \) be a finite set, and let \( f: X \rightarrow \mathbb{R} \) be a function, and let \( c \) be a real number. Then
\[ \sum_{x \in X}cf(x) = c\sum_{x \in X}f(x) \]
8. Monotonicity. Let \( X \) be a finite set, and let \( f: X \rightarrow \mathbb{R} \) and \( g: X \rightarrow \mathbb{R} \) be functions such that \( f(x) \le g(x) \text{ for all } x \in X \). Then
\[ \sum_{x \in X}f(x) \le \sum_{x \in X}g(x) \]
9. Triangle inequality. Let \( X \) be a finite set, and let \( f: X \rightarrow \mathbb{R} \) be a function, then
\[ |\sum_{x \in X}f(x)| \le \sum_{x \in X}|f(x)| \]
BackOnlyText: The 4 sets of series laws: part Ⅱ
Unique to finite sets
The first two properties are trivial and don't appear elsewhere.
Applies to finite sets but not infinite sets (converging series)
Substitution 1 doesn't apply to infinite sets as convergence can't be guaranteed (just imagine the function \( g(i) = 1 \). Substitution 2 is inherent in the definition of infinite set sums, so does not appear as a law. The triangle inequality doesn't apply to infinite sets as their sum is only defined for absolute convergence. I'm not sure why monotonicity isn't defined for infinite sets. converging series, there is no triangle inequality defined as there is no definition (thus no value) for their conditional convergence. Tao doesn't absolutely define a comparison test for absolutely converging series either but I'm not sure why (maybe because it is covered by infinite series?)
Equivalent between finite set and finite series notation
The 2nd substitution rule translates between the two forms, so in a way, part I and part II laws have equivalency.
Context: finite series → finite sets → infinite series → infinite sets (absolutely convergent series)

Title: Fubini's theorem for finite series
Q: Let \( X \) and \( Y \) be finite sets, and let \( f: X \times Y \rightarrow R \) be a function. Then
\[
\begin{aligned}
\sum_{x \in X}\left( \sum_{y \in Y}f(x,y) \right) &= [...] \\
&= [...] \\
&= [...]
\end{aligned}
 \]
A: \sum_{(x,y) \in X \times Y}f(x,y); \sum_{(y,x) \in Y \times X}f(x,y); \sum_{y \in Y} \left( \sum_{x \in X}f(x,y) \right)
BackOnlyText: The proof isn't too involved. Most of the work is done by the lemma:
Lemma
Let \( X \) and \( Y \) be finite sets, and let \( f: X \times Y \rightarrow \mathbb{R} \) be a function. Then
\[ \sum_{x \in X}\left( \sum_{y \in Y}f(x,y) \right) = \sum_{(x,y) \in X \times Y}f(x,y) \]
Proof
Let \( n \) be the number of elements in \( X \). We will induce on \( n \). Let \( P(n) \) be the assertion that the lemma is true for any set \( X \) with \( n \) elements, and any finite set \( Y \) and any function \( f: X \times Y \rightarrow \mathbb{R} \). We wish to prove that \( P(n) \) is true for all natural numbers \( n \).
The base case, \( P(0) \), is easy; it follows from the proposition that the sum over an empty set is zero.
Now suppose that \( P(k) \) is true for some \( k \in \mathbb{N} \); we consider \( P(k+1) \).
Let \( X \) be a set with \( k+1 \) elements. Write \( X \) as \( X = X' \cup \{x_0\} \), where \( x_0 \) is an element of \( X \). Using the 5^th of the 9 propositions on sums over finite sets:
\[\begin{align*}
\sum_{x \in X} \left( \sum_{y \in Y}f(x,y) \right) &= \left( \sum_{x \in X'} \left( \sum_{y \in Y}f(x,y) \right) \right) + \left( \sum_{y \in Y}f(x_0,y) \right) \\
\text{By the induction hypothesis:} \\
&= \sum_{(x,y) \in X' \times Y} f(x,y) + \left( \sum_{y \in Y} f(x_0, y) \right)\\
\text{By the substitution proposition for sums over finite sets:} \\
&= \sum_{(x,y) \in X' \times Y} f(x,y) + \left( \sum_{(x,y) \in \{x_0\} \times Y} f(x, y) \right) \\
\text{by the proposition of sums over disjoint finite sets:} \\
&= \sum_{(x,y) \in X \times Y} f(x,y)\end{align*}\]
Proving Fubini's theorem
The above lemma means we already know that both:
\[ \sum_{x \in X}\left( \sum_{y \in Y}f(x,y) \right) = \sum_{(x,y) \in X \times Y}f(x,y) \]
and
\[ \sum_{y \in Y}\left( \sum_{x \in X}f(x,y) \right) = \sum_{(y,x) \in Y \times X}f(x,y) \]
So it suffices to show that:
\[ \sum_{(x,y) \in X \times Y}f(x,y) = \sum_{(y,x) \in Y \times X}f(x,y)\]
But this follows from the substitution proposition for sums over finite sets by choosing the bijection \( h: X \times Y \rightarrow Y \times X \) defined by \( h(x,y) = (y,x) \).

Title: Infinite series, definition
Q: Tao breaks the definition into two pieces (reasons discussed on flip side). I present the 2^nd part first. See the flip side for the initial definition of a formal series.
Convergence of series, definition
Let \( \sum_{n=m}^{\infty} a_n \) be a formal infinite series. For any integer \( N \ge m \), we define the N^th partial sum \( S_N \) of this series to be
[...]
\( S_N \) is thus a real number.
If the sequence [...] converges to some limit \( L \), then we say that the infinite series \( \sum_{n=m}^{\infty} a_n \) is convergent, and converges to \( L \). We then write
\[ L = \sum_{n=m}^{\infty} a_n \]
and say that \( L \) is the sum of the infinite series \( \sum_{n=m}^{\infty} a_n \). If the sequence of partial sums \( (S_N)_{N=m}^{\infty} \) diverge, then we say that the infinite series \( \sum_{n=m}^{\infty} a_n \) is divergent, and we [...].
A: \[ S_N := \sum_{n=m}^{N} a_n \]; \( (S_N)_{N=m}^{\infty} \); do not assign any real value to the series
BackOnlyText: Formal infinite series, definition
A formal infinite series is any expression of the form
\[ \sum_{n=m}^{\infty}a_n \]
where \( m \) is an integer and \( a_n \) is a real number for any integer \( n \ge m \).
Tao mentions that at this point, this series is only defined formally; in other words, the series hasn't been set to equal anything. [This is all fine, I just don't see why we needed this initial step].
Necessary?
My take is that we could just use a definition like so:
\[ \sum_{n=m}^{\infty}a_n := \lim_{N \rightarrow \infty} \sum_{n=m}^{N}a_n \]
With this definition, the ideas of the infinite sum only taking values when the limit converges is inherited from the definition of the limit. Tao's presentation seems to repeat a lot of these ideas again.
Update on my take: so far at least, limits are only defined in terms of sequences. Tao is explicit about what sequence the limit is operating over. My approach does not make this explicit and is assuming that Tao's step of defining the corresponding sequence occurs.
Tao even continues his use of the term formal series further into the chapter and I haven't seen a clear transition at which the terminology is defined away.
Source: p165-166

Title: Zero tail, and the zero test, propositions
Q: A property of a convergent series is a diminishing tail. This is expressed formally as follows.
Let \( \sum_{n=m}^{\infty}a_n \) be a formal series of real numbers. Then \( \sum_{n=m}^{\infty}a_n \) converges if and only if, for every real number \( \varepsilon > 0 \), there exists an integer \( N \ge m \) such that
[...]
This proposition, by itself, is a little difficult to work with, as computing the partial sums at the tail might not be easy. However, there are a number of corollaries, the first of which is the zero test.
Zero test
Let \( \sum_{n=p}^{\infty}a_n \) be a convergent series of real numbers. Then we must have [...]. In other words, if \( \lim_{n \rightarrow \infty}a_n \) is non-zero or divergent, then the series \( \sum_{n=m}^{\infty}a_n \) is divergent.
A: \[ \left| \sum_{n=p}^{q} a_n \right| \le \varepsilon \text{ for all } p,q \ge N \]; \( \lim_{n \rightarrow \infty}a_n = 0 \)
BackOnlyText: The zero test is an if-then relationship. Convergence of the corresponding sequence to zero is required for a series to be convergent, but the convergence of the sequence to zero is not a sufficient criteria.
Example: The sequence \( a_n := 1 \) is convergent, but does not converge to 0, so the series \( \sum_{n=1}^{\infty}a_n \) must be divergent.
The sequence \( a_n := (-1)^n \) is divergent, so the series \( \sum_{n=1}^{\infty} (-1)^n \) is divergent.
The sequence \( a_n := \frac{1}{n} \) converges to zero; however, the series \( \sum_{n=1}^{\infty} \frac{1}{n} \) is divergent.
Source: p167

Title: Absolute convergence, and the absolute convergence test
Absolute convergence, definition
Let \( \sum_{n=m}^{\infty}a_n \) be a formal series of real numbers. The series is said to be absolutely convergent if the series \( \sum_{n=m}^{\infty}|a_n| \) is convergent.
In order to distinguish convergence from absolute convergence, we sometimes refer to the former as conditional convergence.
Absolute convergence test
Let \( \sum_{n=m}^{\infty}a_n \) be a formal series of real numbers. If the series is absolutely convergent, then it is also conditionally convergent. Furthermore, in this case we have the triangle inequality
\[ \left| \sum_{n=m}^{\infty}a_n \right| \le \sum_{n=m}^{\infty}|a_n| \]
Note that the converse of this proposition is not true; there exists series that are conditionally convergent but are not absolutely convergent.
Source: p167-168

Title: Alternating series test
Q: An alternating series converges if the sequence of elements converges absolutely to zero.
Let \( (a_n)_{n=m}^{\infty} \) be a sequence of real numbers which are non-negative and decreasing, thus \( a_n \ge 0 \) and \( a_n \ge a_{n+1} \) for every \( n \ge m \). Then the series \( \sum_{n=m}^{\infty}(-1)^n a_n \) is convergent if and only if the sequence [...] converges to 0 as \( n \rightarrow \infty \).
A: \( (a_n)_{n=m}^{\infty} \)
BackOnlyText: There is some weight to this proposition, as non-alternating series do not have this property. For example, the series \( \sum_{n=m}^{\infty}\frac{1}{n} \) does not converge. But, \( \sum_{n=m}^{\infty}(-1)^{n}\frac{1}{n} \) in comparison will converge to zero.
Proof outline
The below approach is summarized as:
Let \( N \) be even. For any \( k>0 \), \( S_{N+k} \) is less than \( S_N \) but greater than \( S_{N+1} \). The difference between \( S_N \) and \( S_{N+1} \) approaches zero because the sequence \( (a_n)_{n=m}^{\infty} \) approaches zero. So the sequence of sums \( (S_n)_{n=m}^{\infty} \) is \( \varepsilon \)-steady for every \( \varepsilon > 0 \).
Proof
This proof is crafty, and my attempts at an alternative failed, so there is some value in internalizing the approach.
One side of the iff relationship is easy. From the zero test for series, we know that if \( \sum_{n=m}^{\infty}(-1)^n a_n \) is a convergent series, then the sequence \( ((-1)^n a_n)_{n=m}^{\infty} \) converges to 0, and by the zero test for sequences, we can infer that the sequence (\( a_n)_{n=m}^{\infty} \) must also converge to zero.
Now suppose conversly that (\( (-1)^n a_n)_{n=m}^{\infty} \) converges to 0. For each \( N \ge m \), let \( S_N \) be the partial sum \( S_N := \sum_{n=m}^{N}(-1)a_n \); our job is to show that \( (S_N)_{N=m}^{\infty} \) converges. Observe that
\[
\begin{aligned}
S_{N+2} &= S_N + (-1)^{N+1}a_{N+1} + (-1)^{N+2}a_{N+2} \\
&= S_N + (-1)^{N+1}(a_{N+1} - a_{N+2})
\end{aligned}
\]
By hypothesis \( (a_{N+1} - a_{N+2}) \) is non-negative. Thus we have \( S_{N+2} \ge S_N \), when \( N \) is odd and \( S_{N+2} \le S_N \) when \( N \) is even. Now suppose that \( N \) is even. [What is nice is the the proof proceeds without needing to consider the other case. We arrive at an inequality that works for what we need.]
From the above discussion and induction we see that both:
\( S_{N+2k} \le {S_N} \) and \( S_{N + 2k+1} \ge S_{N+1} = S_N - a_{N+1} \) for all natural numbers \( k \). Also we have \( S_{N + 2k + 1} = S_{N + 2k} - a_{N + 2k + 1} \le S_{N+2k} \). Combining all of this we can say:
\[ S_N - a_{N+1} \le S_{N+2k + 1} \le S_{N+2k} \le S_N \text{ for all k} \]
This means that
\[ S_N - a_{N+1} \le S_n \le S_N \text{ for all } n \ge N \]
This is enough for us to be able to say that \( (S_n)_{n=m}^{\infty} \) is eventually \( a_{N+1} \)-steady. But the sequence \( (a_N)_{N=m}^{\infty} \) converges to 0 as \( N \rightarrow \infty \), which implies that \( (S_n)_{n=m}^{\infty} \) is eventually ε-steady for every \( \varepsilon > 0 \). Thus, \( (S_n)_{n=m}^{\infty} \) converges, and so the series \( \sum_{n=m}^{\infty}(-1)^n a_n \) is convergent.

Title: Telescoping series
Q: Let \( (a_n)_{n=0}^{\infty} \) be a sequence of real numbers which converge to 0, i.e., \( lim_{n \rightarrow \infty} a_n = 0 \). Then the series [...] converges to \( a_0 \).
A: \( \sum_{n=0}^{\infty}(a_n - a_{n+1}) \)
BackOnlyText: A post on The Math Forum talks about the naming of this series:
Source: p170
http://mathforum.org/library/drmath/view/71957.html

Title: The comparison test (and bounded series of non-negative numbers)
We wish to extend the comparison test for finite series to apply to infinite series.
For finite series, the comparison test appeared as a simple opening lemma (7.1.4 f):
\[ \text{When } a_i \le b_i \text{ we have } \sum_{i=m}^{n} a_i \le \sum_{i=m}^{n} b_i \]
For infinite series, we can only make this statement when \( |a_i| \le b_n \), otherwise the sum for \( a \) could diverge. We set this up and prove it below.
First, a useful (and very simple) proposition.
Sums of non-negative numbers are bounded iff they are convergent.
Let \( \sum_{n=m}^{\infty}a_n \) be a formal series of non-negative real numbers. Then this series is convergent if and only if there is a real number \( M \) such that
\[ \sum_{n=m}^{N}a_n \le M \text{ for all integers } N \ge m \]
Proof
Two earlier propositions are:
- Every convergent sequence of real numbers is bounded (6.1.17).
- An increasing sequence which has an upper bound is convergent (6.3.8).
The sequence \( (S_N)_{n=m}^{\infty} \) representing the series \( \sum_{n=m}^{\infty}a_n \) is increasing, by definition. So the above two propositions form each side of the proposition's iff statement.
Now we can introduce the comparison test.
Comparison test
Let \( \sum_{n=m}^{\infty}a_n \) and \( \sum_{n=m}^{\infty}b_n \) be two formal series of real numbers, and suppose that \( |a_n| \le b_n \) for all \( n \ge m \) (thus all numbers in \( (b_n) \) must be non-negative). Then if \( \sum_{n=m}^{\infty}b_n \) is convergent, then \( \sum_{n=m}^{\infty}a_n \) is absolutely convergent, and in fact
\[  \left| \sum_{n=m}^{\infty}a_n \right| \le \sum_{n=m}^{\infty}|a_n| \le  \sum_{n=m}^{\infty}b_n \]
BackOnlyText: Proof
Let \( SA_N := \sum_{n=m}^{N}|a_n|  \) be the Nth partial sum of \(  \sum_{n=m}^{\infty}|a_n| \) and \( SB_N := \sum_{n=m}^{N}b_n \) be the Nth partial sum of \( \sum_{n=m}^{\infty}b_n \). As \( \sum_{n=m}^{\infty}b_n \) is bounded according to the proposition above, and by the the comparison test for finite series we have: \[ SA_N \le SB_N \le M \text{ for all } N \ge m \text{ and for some bounding real } M \] Through this transitivity, \( (SA_N)_{N=n}^{\infty} \) also meets the conditions for being bounded. Then from the above propositive again, it must be that \( \sum_{n=m}^{\infty}|a_n| \) is convergent. The Absolute Convergence Test informs us that absolute convergence implies conditional convergence, and in addition that \( \left| \sum_{n=m}^{\infty}a_n \right| \le \sum_{n=m}^{\infty}|a_n| \). Thus, with all of the below infinite series converging we can compare them like so:
\[ \left| \sum_{n=m}^{\infty}a_n \right| \le \sum_{n=m}^{\infty}|a_n| \le \sum_{n=m}^{\infty}b_n \]
We can also run the Comparison Test in the contrapositive:
If \( \sum_{n=m}^{\infty}a_n \) is not absolutely convergent then \( \sum_{n=m}^{\infty}b_n \) is not conditionally convergent.
Source: p171

Title: Geometric series
Q: Let \( x \) be a real number. If \( |x| \ge 1 \), then the series \( \sum_{n=0}^{\infty} x^n \) is divergent. If however \( |x| \le 1 \), then the series is absolutely convergent and the sum is given by:
[...]
On the reverse side, there is some geometric visual reasoning; can you remember what it looks like?
A: \[ \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \]
BackOnlyText: Geometric interpretation
The name "geometric" might have root in whatever led to the geometric mean getting its name. Anayway, below is some nice geometric reasoning. In particular, it shows why when \(r \ge 1\) the series will not converge.
To grasp the reasoning, note that with a slope of \(x\), a movement of \(1 \) horizontally will lead to vertical movement of \(x\), and horizontal movement of \(x\) will lead to vertical movement of \(x^2\).
Standard proof
The proof of this is is famous and well known
Todo: sort out html video embedding.
Todo: it would be good to add a proof that addresses all points in the definition/proposition in the style of the book.

Title: Cauchy criterion, harmonic series and the Riemann-zeta function
Let \( (a_n)_{n=1}^{\infty} \) be a decreasing sequence of non-negative real numbers. Then the series \( \sum_{n=1}^{\infty}a_n \) is convergent if and only if the series
\[ \sum_{k=0}^{\infty} 2^ka_{2^k} = a_1 + 2a_2 + 4a_4 + 8a_8 + ... \]
is convergent. This is the Cauchy criterion.
Harmonic series
The Cauchy criterion can be used to show that the Harmonic series, \( \sum_{n=1}^{\infty}\frac{1}{n} \), is divergent. Yet \( \sum_{n=1}^{\infty}\frac{1}{n^2} \) is convergent.
Riemann-zeta function
The quantity \( \sum_{n=1}^{\infty}\frac{1}{q^n} \) is called the Riemann-zeta function of q and is denoted by \( \zeta(q) \). This function is very important in number theory in particular for investigating the distribution of primes. There is a famous unsolved problem regarding this function called the Riemann hypothesis.
BackOnlyText: An interesting feature of this criterion is that it only uses a small number of elements of the sequence \( a_n \) (namely, those elements whose index is a power of 2, \( n = 2_k \text{ for some integer k > 0} \)) in order to deterimen whether the whole series is convergent or not.
Intuition
There is a visual/geometric understanding that makes the mechanism of the proof clear. See the drawing below.
The Cauchy Criterion introduces the idea of bunching together segments of the sum to make a new sum, and by clever bunching we can see if the series is decreasing fast enough for the sum to converge.
I think it is correct to say that if a series diverges, then there will exist a bunching of the sum such that each bunch is greater than some constant. This idea can be seen a bit more clearly if you try and create a clever bunching for the sum \( \sum_{n=1}^{\infty}\frac{1}{n^2} \)—you will find that there is no bunching that will work to form a constant, instead the bunches will keep getting smaller.
Proof
The proof of this is not short. The book has a more consise approach than what I took. The first half of my approach is:
The choice of \( 2^k \) seems quite natural when tackling the harmonic series. Try to group the harmonic series elements as above such that they sum to 1. You end up with \( 1\times \frac{1}{1}, 2\times \frac{1}{2}, 4\times \frac{1}{4}... \). Already, this coves \(1+2+4 = 7\) elements, so the next will be the 8th whose value is \( \frac{1}{8} \) requiring a grouping of 8 elements, bringing us up to 15, and so on.
Example: Divergence of the harmonic series
The proof of the following proposition is an example of the application of the Cauchy criterion.
Let \( q > 0 \) be a rational number. Then the series \( \sum_{n=1}^{\infty} \frac{1}{n^q} \) is convergent when \( q > 1 \) and divergent when \( q \le 1 \).
A particular divergent case of this series is the harmonic series: \( \sum_{n=1}^{\infty} \frac{1}{n} \). In comparison, \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is convergent.
Proof
The sequence \( (\frac{1}{n^q})_{n=1}^{\infty} \) is non-negative and decreasing (by Lemma 5.6.9 d), and so the Cauchy criterion applies. Thus, the sequence is convergent if and only if
\[ \sum_{k=0}^{\infty}2^k \frac{1}{(2^k)^q} \]
is convergent. By the laws of exponentiation we can rewrite this as
\[ \sum_{k=0}^{\infty}(2^{1-q})^k \]
This is a geometric series (\( \sum_{k=0}^{\infty}x^k \)) and converges if and only if \( |x| < 1 \). Thus our original \( \sum_{n=1}^{\infty} \frac{1}{n^q} \) will converge if and only if \( |2^{1-q}| < 1 \), which happens if and only if \( q > 1 \) (why?).
Source: p174

Title: Rearrangement of infinite series
Q: A feature of finite series which we will recap here is that any rearrangement of the terms of the series does not affect the sum. For example:
\[ a_1 + a_2 + a_3 + a_4 = a_4 + a_1 + a_3 + a_2 \]
This comes from the first property of substitution:
If \( X \) is a finite set, \( f: X \rightarrow R \) is a function, and \( g : Y \rightarrow X \) is a bijection, then:
[...]
If we consider any bijection \( g \) from-to the same set \( \{ i \in \mathbb{Z} : n \le i \le m \} \), then we can say:
\[ \sum_{i=n}^{m} a_i = \sum_{i=n}^{m} a_{g(i)} \]
which is the basis for the rearrangement example above.
Can we rearrange the terms of an infinite series and get the same result? Yes and no.
- An absolutely convergent series: [...]
- Conditionally, but not absolutely convergent series: [...]
A: \[ \sum_{x \in X}f(x) = \sum_{y \in Y}f(g(y)) \]; any rearrangement of terms produces the same sum.; the terms can be arranged to make the series converge to any value, or to diverge.
BackOnlyText: In summary: rearrangement of absolute convergent series is safe, rearrangement of a non-absolutely convergent series is risky. The latter is due to being able to manipulate the occurances of positives and negatives which, in the original series, are balancing each other out in order to achieve convergence.
The full proposition for absolutely convergent series is as follows.
Rearrangement of infinite series
Let \( \sum_{n=0}^{\infty}a_n \) be an absolutely convergent series of real numbers and let \( f: N \rightarrow N \) be a bijection. Then \( \sum_{m=0}^{\infty} a_{f(m)} \) is also absolutely convergent, and has the same sum:
\[ \sum_{n=0}^{\infty} a_n = \sum_{n=0}^{\infty} a_{f(m)} \]
Proof
Todo

Title: Ratio test
Q: Let \( \sum_{n=m}^{\infty}a_n \) be a series of non-zero numbers.
- If [...], then the series \( \sum_{n=m}^{\infty}a_n \) is absolutely convergent (hence conditionally convergent).
- If [...], then the series \( \sum_{n=m}^{\infty}a_n \) is not conditionally convergent (hence not absolutely convergent).
- Otherwise, we cannot assert any conclusion.
The non-zero requirement is needed to avoid a division by zero.
A: \( \limsup_{n \rightarrow \infty} \frac{|a_{n+1}| }{|a_n|} < 1 \); \( \limsup_{n \rightarrow \infty} \frac{|a_{n+1}| }{|a_n|} > 1 \)
BackOnlyText: Intuition
Tao proves this result using the Root Test along with a lemma. However, for understanding and remembering the Ratio Test, I think it is best described in terms of the behaviour of the geometric series.
For a geometric series, \( \sum_{n=m}^{\infty} a_n = \sum_{n=m}^{\infty} r^{n} = r^m + r^{m+1} + r^{m+2} ... \), we know that it converges absolutely if \( r < 1 \) and diverges otherwise.
This criteria can be rephrased using the ratio between the terms: \( \frac{a_{n+1}}{a_n} = r < 1 \) implies that the series converges (and diverges otherwise). If we extend this intuition to all series, we have the ratio test.
Source: p181
https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-8/v/ratio-test-convergence

Title: The Root Test
Q: Let \( \sum_{n=m}^{\infty}a_n \) be a series of real numbers and let [\( \alpha = ? \)].
1. If \( \alpha < 1 \), then the series \( \sum_{n=m}^{\infty}a_n \) is absolutely convergent (and hence conditionally convergent).
2. If \( \alpha > 1 \), then the series \( \sum_{n=m}^{\infty}a_n \) is not conditionally convergent (and hence is not absolutely convergent either).
3. If \( \alpha = 1 \), this test does not assert any conclusion.
The famous Root Test.
A: \( \alpha = \limsup_{n \rightarrow \infty}|a_n|^{\frac{1}{n} } \)
BackOnlyText: Intuition
The root test applies the same principle used for the ratio test: that of comparing with a geometric series.
If \( |a_n|^{\frac{1}{n}} < 1 \) for all \( n \) greater than some \( N \), then we have \( |a_n| < \alpha^n \text{ for some } \alpha < 1 \). We know that the geometric series \( \sum_{n=N}^{\infty}\alpha^n \) converges, so by the comparison principle, then so too does the series \( \sum_{n=m}^{\infty}a_n \).
Proof
The proof follows the same lines as the intuition; added details cover the behaviour of the limit supremum.
First suppose that \( \alpha < 1 \).
- We can find a \( \varepsilon > 0 \) such that \( \alpha + \varepsilon < 1 \).
- There must exist an \( N \ge m \) such that \( |a_n|^{\frac{1}{n}} < \alpha + \varepsilon \) (By Prop 6.4.12(a): terms are eventually all less than any number larger than the limit supremum).
- Manipulating the exponent we can write: \( |a_n| < (\alpha + \varepsilon)^n \) for all \( n \ge N \).
- The geometric series \( \sum_{n=m}^{\infty} (\alpha + \varepsilon)^n \) converges as \( (\alpha + \varepsilon) < 1 \).
- By the comparison principle, \( \sum_{n=m}^{\infty} |a_n| \) must also converge.
- Thus, \( \sum_{n=m}^{\infty} a_n \) is absolutely convergent.
Now suppose that \( \alpha > 1 \).
- For any \( N \ge m \) there exists a \( k \) such that \( |a_k|^{\frac{1}{k}} > 1 \) (By Prop 6.4.12(b): there are always some terms greater than any number lower than the limit supremum).
- For such \( k \), we also have \( |a_k| > 1 \).
- Thus, \( (a_n)_{n=m}^{\infty} \) is not eventually \( \varepsilon \)-close for all \( \varepsilon > 0 \) (e.g. 1-close, 0.5-close).
- By the Zero Test we can conclude that \( \sum_{n=m}^{\infty} a_n \) is not conditionally convergent.
Source: p179

Title: Series laws Ⅰ: finite series laws
1. Continuation. Let \( m \le n < p \) be integers, and let \( a_i \) be a real number assigned to each integer \( m \le i \le p \). The we have
\[ \sum_{i=m}^{n} a_i + \sum_{i=n+1}^{p} a_i = \sum_{i=m}^{p}a_i \]
2. Indexing shift. Let \( m \le n \) be integers, \( k \) be another integer, and let \( a_i \) be a real number assigned to each integer \( m \le i \le n \). Then we have
\[ \sum_{i=m}^{n}a_i = \sum_{j=m+k}^{n+k}a_{j-k}  \]
This one above, I actually had trouble proving. It is almost too obvious.
3. Linearity part I. Let \( m \le n \) be integers, and let \( a_i, b_i \) be real numbers assigned to each integer \( m \le i \le n \). Then we have
\[ \sum_{i=m}^{n}(a_i + b_i) = \left( \sum_{i=m}^{n}a_i \right) + \left( \sum_{i=m}^{n}b_i \right) \]
4. Linearity part II. Let \( m \le n \) be integers, and let \( a_i \) be a real number assigned to each integer \( m \le i \le n \), and let \( c \) be another real number. Then we have
\[ \sum_{i=m}^{n}(c a_i) = c \left( \sum_{i=m}^{n}a_i \right) \]
5. Triangle inequality. Let \( m \le n \) be integers, and let \( a_i \) be a real number assigned to each integer \( m \le i \le n \). Then we have
\[ \left| \sum_{i=m}^{n}a_i \right| \le \sum_{i=m}^{n}|a_i| \]
6. Comparison test. Let \( m \le n \) be integers, and let \( a_i, b_i \) be real numbers assigned to each integer \( m \le i \le n \). Suppose that \( a_i \le b_i \) for all \( m \le i \le n \). Then we have
\[ \sum_{i=m}^{n}a_i \le \sum_{i=m}^{n}b_i \]
finite series → finite sets → infinite series → infinite sets (absolutely convergent series)
BackOnlyText: The 4 sets of series laws: part Ⅰ
Unique to finite series
The first two properties of finite series do not have parallels in any of the other 3 types of series. All other properties reappear for finite sets and infinite series.
Missing from absolutly converging series
For absolutely converging series, there is no triangle inequality defined as there is no definition (thus no value) for their conditional convergence. Tao doesn't define a comparison test for absolutely converging series either but I'm not sure why (maybe because it is covered by infinite series?)
Source: p157-158