07 Series Cauchy Criterion, Absolute and Conditional Convergence, Riemann Rearrangement Theorem, Dirichlet's Theorem

Lecture from: 11.03.2024 | Video: Video ETHZ

This lecture continues our exploration of series, focusing on the Cauchy Criterion for series convergence, properties of series with non-negative terms, the crucial concept of absolute convergence, and the intriguing behavior of conditionally convergent series under rearrangement.

Clicker Question: Interval Bisection and Series Representation

Constructing a Number via Nested Intervals

Let’s consider a nested interval construction. We start with $I_1=[0,1]$. We construct a sequence of nested intervals $I_1\supseteq I_2\supseteq I_3\supseteq \dots$ by alternately choosing the right and left half of the preceding interval.

Specifically, for $n\ge 1$:

• If $n$ is odd, $I_{n+1}$ is the closed right half of $I_n$.
• If $n$ is even, $I_{n+1}$ is the closed left half of $I_n$.

We want to determine the value $c\in \mathbb{R}$ such that ${\bigcap }_{n=1}^{\infty }{I}_n=\{c\}$.

Visualizing the Interval Bisection

Imagine starting with $[0,1]$.

1. $I_1=[0,1]$
2. $I_2=[\frac{1}{2},1]$ (right half of $I_1$)
3. $I_3=[\frac{1}{2},\frac{3}{4}]$ (left half of $I_2$)
4. $I_4=[\frac{5}{8},\frac{3}{4}]$ (right half of $I_3$)
5. $I_5=[\frac{5}{8},\frac{11}{16}]$ (left half of $I_4$) … and so on.

We can see that the intervals are nested and their lengths are halving at each step. By the Nested Interval Theorem, their intersection is a single point $c$.

Calculating the Value $c$ as a Series

Let’s express $c$ as a series. We can track the left endpoint of each interval.

• Start at 0.
• Move right by $\frac{1}{2}$ to get to the start of $I_2$.
• Move neither left nor right for $I_3$ (left endpoint of $I_3$ is the same as $I_2$).
• Move right by $\frac{1}{8}$ to get to the start of $I_4$.
• Move neither left nor right for $I_5$ (left endpoint of $I_5$ is the same as $I_4$).
• Move right by $\frac{1}{32}$ to get to the start of $I_6$. …

So, we can express $c$ as a sum:

$c=\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{128}{1}+\cdots =\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\dots$

This is a geometric series. We can factor out $\frac{1}{2}$:

$c=\frac{1}{2}\left(1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dots \right)=\frac{1}{2}\left(1+\frac{1}{4}+(\frac{1}{4}{)}^2+(\frac{1}{4}{)}^3+\dots \right)$

This is a geometric series with first term $a=1$ and common ratio $q=\frac{1}{4}$. Since $|q|=\frac{1}{4}<1$, the series converges to $\frac{a}{1-q}=\frac{1}{1-\frac{1}{4}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$.

Therefore,

$c=\frac{1}{2}\cdot \frac{4}{3}=\frac{2}{3}$

Alternatively, we can directly use the geometric series formula with first term $a=\frac{1}{2}$ and common ratio $q=\frac{1}{4}$:

$c={\sum }_{k=0}^{\infty }\frac{1}{2}\cdot (\frac{1}{4}{)}^k=\frac{\frac{1}{2}}{1-\frac{1}{4}}=\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{1}{2}\cdot \frac{4}{3}=\frac{2}{3}$

Thus, the intersection of these nested intervals is the single point $\{\frac{2}{3}\}$.

Cauchy Criterion for Series Convergence

Theorem Statement

The Cauchy Criterion provides a way to determine if a series converges without needing to know its limit in advance, similar to its application for sequences.

Theorem (Cauchy Criterion for Series): The series ${\sum }_{k=1}^{\infty }{a}_k$ is convergent if and only if for every $ϵ>0$, there exists $N\in \mathbb{N}$ such that for all $m>n>N$,

$\left|{\sum }_{k=n+1}^m{a}_k\right|=|a_{n+1}+a_{n+2}+\cdots +a_m|<ϵ$

Connection to Partial Sums

Let $S_n={\sum }_{k=1}^n{a}_k$ be the sequence of partial sums. Then for $m>n$,

$S_m-S_n=\left({\sum }_{k=1}^m{a}_k\right)-\left({\sum }_{k=1}^n{a}_k\right)={\sum }_{k=n+1}^m{a}_k=a_{n+1}+a_{n+2}+\cdots +a_m$

Thus, the Cauchy Criterion for series is directly derived from the Cauchy Criterion for the sequence of partial sums $(S_n)$. The series converges if and only if $(S_n)$ is a Cauchy sequence, which is equivalent to the condition stated in the theorem.

Non-Negative Series and Bounded Partial Sums

Series with Non-Negative Terms

Consider a series ${\sum }_{k=1}^{\infty }{a}_k$ where $a_k\ge 0$ for all $k\ge 1$. In this case, the sequence of partial sums $(S_n{)}_{n\ge 1}$ is monotonically increasing (non-decreasing) because $S_{n+1}=S_n+a_{n+1}\ge S_n$.

Convergence Condition for Non-Negative Series

Theorem: Let ${\sum }_{k=1}^{\infty }{a}_k$ be a series with $a_k\ge 0$ for all $k\ge 1$. Then the series converges if and only if the sequence of partial sums $(S_n{)}_{n\ge 1}$ is bounded above.

Proof:

• ($\Rightarrow$) If ${\sum }_{k=1}^{\infty }{a}_k$ converges, then $(S_n)$ is bounded. This is true for any convergent sequence of partial sums, as convergent sequences are always bounded.

• ($\Leftarrow$) If $(S_n)$ is bounded above, then ${\sum }_{k=1}^{\infty }{a}_k$ converges. Since $(S_n)$ is monotonically increasing and bounded above, by the Monotone Convergence Theorem (Weierstrass Theorem), the sequence $(S_n)$ converges. Therefore, the series ${\sum }_{k=1}^{\infty }{a}_k$ converges.

Remark on Possibilities for Non-Negative Series

For a series ${\sum }_{k=1}^{\infty }{a}_k$ with $a_k\ge 0$, there are only two possibilities:

1. Bounded Partial Sums: The partial sums $(S_n)$ are bounded above. In this case, the series converges to a finite value, and we write ${\sum }_{k=1}^{\infty }{a}_k<\infty$.
2. Unbounded Partial Sums: The partial sums $(S_n)$ are unbounded above. Since $(S_n)$ is monotonically increasing, it must diverge to infinity. In this case, we write ${\sum }_{k=1}^{\infty }{a}_k=\infty$.

Corollary: Comparison Test (Vergleichssatz)

Corollary (Comparison Test): Let ${\sum }_{k=1}^{\infty }{a}_k$ and ${\sum }_{k=1}^{\infty }{b}_k$ be series with $0\le a_k\le b_k$ for all $k\ge 1$. Then:

1. If ${\sum }_{k=1}^{\infty }{b}_k$ converges, then ${\sum }_{k=1}^{\infty }{a}_k$ converges.
2. If ${\sum }_{k=1}^{\infty }{a}_k$ diverges, then ${\sum }_{k=1}^{\infty }{b}_k$ diverges.

Proof:

1. If ${\sum }_{k=1}^{\infty }{b}_k$ converges, the partial sums of $\sum b_k$ are bounded above. Since $0\le a_k\le b_k$, the partial sums of $\sum a_k$ are also bounded above by the same bound (or even a smaller bound). As $\sum a_k$ is a non-negative series with bounded partial sums, it converges.

2. This is the contrapositive of statement 1. If ${\sum }_{k=1}^{\infty }{a}_k$ diverges, its partial sums are unbounded. Since $0\le a_k\le b_k$, the partial sums of $\sum b_k$ must also be unbounded (at least as large), so ${\sum }_{k=1}^{\infty }{b}_k$ also diverges.

Example: Partial Fraction Decomposition and Convergence

Consider the series ${\sum }_{k=2}^{\infty }\frac{1}{k^2-k}={\sum }_{k=2}^{\infty }\frac{1}{k(k-1)}$.

We can use partial fraction decomposition:

$\frac{1}{k(k-1)}=\frac{A}{k}+\frac{B}{k-1}=\frac{A(k-1)+Bk}{k(k-1)}=\frac{(A+B)k-A}{k(k-1)}$

Comparing coefficients:

• $A+B=0$
• $-A=1⟹A=-1$
• $B=-A=1$

So, $\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}$.

The series becomes a telescoping series:

${\sum }_{k=2}^{\infty }\left(\frac{1}{k-1}-\frac{1}{k}\right)=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots$

The partial sum $S_n={\sum }_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=1-\frac{1}{n}$.

As $n\to \infty$, $S_n\to 1$. Therefore, ${\sum }_{k=2}^{\infty }\frac{1}{k^2-k}=1$.

Comparison Test Application

For $k\ge 2$, we have $k^2-k=k(k-1)\ge (k-1)(k-1)=(k-1{)}^2$. Thus, $\frac{1}{k^2-k}\le \frac{1}{(k-1{)}^2}$.

We know that ${\sum }_{k=1}^{\infty }\frac{1}{k^2}$ converges (p-series with $p=2>1$). This suggests that ${\sum }_{k=2}^{\infty }\frac{1}{k^2-k}$ might also converge by comparison. However, $\frac{1}{k^2-k}\le \frac{1}{(k-1{)}^2}$ is not directly helpful to show convergence, as we would need a larger convergent series to bound it from above in a direct comparison.

Instead, consider $k^2-k\ge \frac{1}{2}{k}^2$ for sufficiently large $k$ (e.g., $k\ge 2$). Then $\frac{1}{k^2-k}\le \frac{2}{k^2}$. Since ${\sum }_{k=1}^{\infty }\frac{2}{k^2}=2{\sum }_{k=1}^{\infty }\frac{1}{k^2}$ converges, by the comparison test, ${\sum }_{k=2}^{\infty }\frac{1}{k^2-k}$ also converges.

Absolute Convergence

Definition of Absolute Convergence

Definition: A series ${\sum }_{k=1}^{\infty }{a}_k$ is called absolutely convergent if the series of absolute values ${\sum }_{k=1}^{\infty }|a_k|$ is convergent.

Theorem: Absolute Convergence Implies Convergence

Theorem: If a series ${\sum }_{k=1}^{\infty }{a}_k$ is absolutely convergent, then it is also convergent. Furthermore, the absolute value of the sum is less than or equal to the sum of the absolute values:

$\left|{\sum }_{k=1}^{\infty }{a}_k\right|\le {\sum }_{k=1}^{\infty }|a_k|$

This is sometimes referred to as the “Triangle Inequality for Series.”

Proof (using Cauchy Criterion):

Since ${\sum }_{k=1}^{\infty }|a_k|$ is convergent, by the Cauchy Criterion for series, for every $ϵ>0$, there exists $N\in \mathbb{N}$ such that for all $m>n>N$,

${\sum }_{k=n+1}^m|a_k|=|a_{n+1}|+|a_{n+2}|+\cdots +|a_m|<ϵ$

Now consider the partial sums of ${\sum }_{k=1}^{\infty }{a}_k$. For $m>n>N$,

$\left|{\sum }_{k=n+1}^m{a}_k\right|=|a_{n+1}+a_{n+2}+\cdots +a_m|\le |a_{n+1}|+|a_{n+2}|+\cdots +|a_m|={\sum }_{k=n+1}^m|a_k|<ϵ$

The inequality used here is the triangle inequality for a finite sum.

Since $\left|{\sum }_{k=n+1}^m{a}_k\right|<ϵ$ for all $m>n>N$, by the Cauchy Criterion for series, the series ${\sum }_{k=1}^{\infty }{a}_k$ is convergent.

To show the inequality for the sums, consider the partial sums $S_n={\sum }_{k=1}^n{a}_k$ and $T_n={\sum }_{k=1}^n|a_k|$. We know that $|S_n|=|{\sum }_{k=1}^n{a}_k|\le {\sum }_{k=1}^n|a_k|=T_n$ (triangle inequality). Since ${\sum }_{k=1}^{\infty }{a}_k$ converges to some value $S$ and ${\sum }_{k=1}^{\infty }|a_k|$ converges to some value $T$, taking the limit as $n\to \infty$ in the inequality $|S_n|\le T_n$ gives $|S|\le T$, which is $\left|{\sum }_{k=1}^{\infty }{a}_k\right|\le {\sum }_{k=1}^{\infty }|a_k|$.

Example: Conditional Convergence

Example: Alternating Harmonic Series

Consider the alternating harmonic series: ${\sum }_{k=1}^{\infty }\frac{(-1{)}^{k+1}}{k}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$.

• Absolute Convergence? No, the series is not absolutely convergent because ${\sum }_{k=1}^{\infty }\left|\frac{(-1{)}^{k+1}}{k}\right|={\sum }_{k=1}^{\infty }\frac{1}{k}$ is the divergent harmonic series.

• Convergence? Yes, the alternating harmonic series is convergent. We will prove this using the Leibniz Criterion.

Leibniz Criterion (Alternating Series Test)

Theorem (Leibniz Criterion): Let $(a_k{)}_{k\ge 1}$ be a sequence of real numbers such that:

1. $(a_k{)}_{k\ge 1}$ is monotonically decreasing ($a_{k+1}\le a_k$ for all $k\ge 1$).
2. $a_k\ge 0$ for all $k\ge 1$.
3. ${\lim }_{k\to \infty }{a}_k=0$.

Then the alternating series ${\sum }_{k=1}^{\infty }(-1{)}^{k+1}{a}_k$ converges. Furthermore, the sum $S$ and the partial sums $S_n$ satisfy the bound on the remainder:

$|S-S_n|\le a_{n+1}$ and the sum $S$ lies between any two consecutive partial sums: $a_1-a_2=S_2\le S\le S_1=a_1$, and in general, for all $n\ge 1$, $S_{2n}\le S\le S_{2n-1}$.

Proof Sketch:

Let $S_n={\sum }_{k=1}^n(-1{)}^{k+1}{a}_k$ be the partial sums.

• Subsequence of even partial sums $(S_{2n}{)}_{n\ge 1}$ is monotonically increasing and bounded above. $S_{2n+2}-S_{2n}=a_{2n+1}-a_{2n+2}\ge 0$ (since $a_{2n+1}\ge a_{2n+2}$). Also, $S_{2n}=(a_1-a_2)+(a_3-a_4)+\cdots +(a_{2n-1}-a_{2n})=a_1-(a_2-a_3)-\cdots -a_{2n}\le a_1$.

• Subsequence of odd partial sums $(S_{2n-1}{)}_{n\ge 1}$ is monotonically decreasing and bounded below. $S_{2n+1}-S_{2n-1}=a_{2n+1}-a_{2n}\le 0$ (since $a_{2n}\ge a_{2n+1}$). Also, $S_{2n-1}=S_{2n}+a_{2n-1}\ge S_{2n}\ge S_2=a_1-a_2$.

• Both subsequences converge to the same limit. ${\lim }_{n\to \infty }{S}_{2n}=L_1$ and ${\lim }_{n\to \infty }{S}_{2n-1}=L_2$. But $S_{2n-1}-S_{2n}=a_{2n}\to 0$ as $n\to \infty$. So $L_2-L_1=0$, and $L_1=L_2=L$. Since both even and odd subsequences converge to the same limit, the entire sequence $(S_n)$ converges to $L$.

• Error Bound: The difference between consecutive partial sums is $|S_n-S_{n-1}|=|a_n|=a_n$. The error $|S-S_n|$ is bounded by the magnitude of the first omitted term $a_{n+1}$.

Example: Applying Leibniz Criterion to Alternating Harmonic Series

For the alternating harmonic series, $a_k=\frac{1}{k}$.

1. $(a_k)$ is monotonically decreasing.
2. $a_k\ge 0$.
3. ${\lim }_{k\to \infty }{a}_k=0$.

Thus, by the Leibniz Criterion, the alternating harmonic series converges.

Conditional Convergence and Rearrangements

Definition of Conditional Convergence

A series ${\sum }_{k=1}^{\infty }{a}_k$ is called conditionally convergent if it is convergent, but not absolutely convergent.

The alternating harmonic series is an example of a conditionally convergent series.

Rearrangements of Series

A series ${\sum }_{k=1}^{\infty }{a}_k^{\prime }$ is a rearrangement of the series ${\sum }_{k=1}^{\infty }{a}_k$ if there is a bijection $\sigma :{\mathbb{N}}^{\ast }\to {\mathbb{N}}^{\ast }$ such that $a_k^{\prime }=a_{\sigma (k)}$ for all $k\in {\mathbb{N}}^{\ast }$. Essentially, a rearrangement is just reordering the terms of the series.

Riemann Rearrangement Theorem

Riemann Rearrangement Theorem: Let ${\sum }_{k=1}^{\infty }{a}_k$ be a conditionally convergent series of real numbers. Then for any real number $A\in \mathbb{R}$, there exists a rearrangement ${\sum }_{k=1}^{\infty }{a}_k^{\prime }$ of ${\sum }_{k=1}^{\infty }{a}_k$ such that ${\sum }_{k=1}^{\infty }{a}_k^{\prime }=A$. Furthermore, there exist rearrangements that diverge to $\infty$ or $-\infty$.

Yes, you actually read that right…

This theorem highlights the drastically different behavior of conditionally convergent series compared to absolutely convergent series. Rearranging a conditionally convergent series can change its sum to any desired value, or even make it diverge!

Example: Rearranging the Alternating Harmonic Series

Consider the alternating harmonic series: $S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots =\log 2\approx 0.693$.

Let’s rearrange it to get a sum of $\frac{1}{2}\log 2\approx 0.3465$. We take one positive term, then two negative terms, then one positive, then two negative, and so on:

$S^{\prime }=1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\frac{1}{10}-\frac{1}{12}+\dots$

We can group terms:

$S^{\prime }=\left(1-\frac{1}{2}\right)-\frac{1}{4}+\left(\frac{1}{3}-\frac{1}{6}\right)-\frac{1}{8}+\left(\frac{1}{5}-\frac{1}{10}\right)-\frac{1}{12}+\dots$ $S^{\prime }=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots =\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\dots \right)=\frac{1}{2}S=\frac{1}{2}\log 2$

The sum changes! This demonstrates that rearrangements of conditionally convergent series can lead to different sums.

Theorem: Dirichlet’s Theorem on Rearrangements of Absolutely Convergent Series

In contrast to conditionally convergent series, absolutely convergent series behave much more predictably under rearrangement.

Theorem (Dirichlet’s Theorem): If ${\sum }_{k=1}^{\infty }{a}_k$ is absolutely convergent, then every rearrangement ${\sum }_{k=1}^{\infty }{a}_k^{\prime }$ of ${\sum }_{k=1}^{\infty }{a}_k$ is also absolutely convergent, and every rearrangement has the same sum.

This theorem highlights the robustness of absolute convergence. For absolutely convergent series, the order of summation does not matter; you can rearrange the terms without changing the sum or losing absolute convergence.

This lecture has explored several advanced aspects of series, emphasizing the Cauchy Criterion, the distinction between absolute and conditional convergence, and the surprising consequences of rearranging conditionally convergent series. These concepts are crucial for a deeper understanding of infinite sums and their properties in real and complex analysis.

Continue here: 08 More Convergence Criteria, Dirichlet’s Theorem, Ratio, Root Tests, and Power Series