# Cauchy criterion – Serlo

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In the section about limits we have seen that the Cauchy criterion provides an alternative characterization of convergence. A sequence is convergent if and only if is a Cauchy sequence. The convergence of a series is defined over the convergence of the sequence of its partial sums. Since the convergence of series traces back to the convergence of sequences, we can also use the Cauchy criterion for series, and that way prove the convergence or divergence of a series.

The Cauchy criterion is named after the French mathematician Augustin Louis Cauchy, because he was the first to introduce this convergence criterion in his textbook „Cours d'Analyse“ (1821). [1].

## Derivation of the Cauchy criterion

### Repetition of required terminology

Cauchy sequences are sequences, where the members will get arbitrary close to each other. If we have a Cauchy sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ for every maximum distance ${\displaystyle \epsilon >0}$ there is a minimal index ${\displaystyle N\in \mathbb {N} }$, so that starting from that index ${\displaystyle a_{N}}$ for two later members ${\displaystyle a_{n}}$ and ${\displaystyle a_{m}}$ the distance ${\displaystyle |a_{n}-a_{m}|}$ is smaller than ${\displaystyle \epsilon }$. Thus for a Cauchy sequence we have:

${\displaystyle \forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n,m\geq N:|a_{n}-a_{m}|<\epsilon }$

For the derivation we also need to define the convergence of series: A series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is convergent if and only if the sequence of its partial sums

${\displaystyle \left(\sum _{k=1}^{n}a_{k}\right)_{n\in \mathbb {N} }=\left(\sum _{k=1}^{1}a_{k},\quad \sum _{k=1}^{2}a_{k},\quad \sum _{k=1}^{3}a_{k},\quad \sum _{k=1}^{4}a_{k},\ \ldots \right)}$

is convergent.

### Derivation

Let ${\displaystyle S_{n}}$ be te ${\displaystyle n}$-th partial sum, so the sum of the first ${\displaystyle n}$ summands:

${\displaystyle S_{n}=a_{1}+a_{2}+a_{3}+\ldots +a_{n}=\sum _{k=1}^{n}a_{k}}$

We assume that the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is convergent. From definition the sequence ${\displaystyle (S_{n})_{n\in \mathbb {N} }}$ converges, therefore it also satisfies the Cauchy criterion. We can plug ${\displaystyle (S_{n})_{n\in \mathbb {N} }}$ into the above Cauchy criterion for sequences:

${\displaystyle \forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n,m\geq N:|S_{n}-S_{m}|<\epsilon }$

The distance ${\displaystyle |S_{n}-S_{m}|}$ becomes arbitrary small, and this part of the formula can be further broken down. Assume that ${\displaystyle n>m}$. Then:

{\displaystyle {\begin{aligned}|S_{n}-S_{m}|&=\left|\sum _{k=1}^{n}a_{k}-\sum _{l=1}^{m}a_{l}\right|\\&=|({\color {OliveGreen}a_{1}+\ldots +a_{m}}+a_{m+1}+\ldots a_{n})-({\color {OliveGreen}a_{1}+\ldots +a_{m}})|\\&=|a_{m+1}+a_{m+2}+\ldots +a_{n}|\\&=\left|\sum _{k=m+1}^{n}a_{k}\right|\\&=\left|\sum _{k=1}^{n-m}a_{m+k}\right|\end{aligned}}}

We observe: If a series is convergent, then the sum of the subsequent summands with arbitrary but fixed length will get arbitrary small with increasing start-index. Thus with the convergence of a series we have:

${\displaystyle \forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n>m\geq N:\left|\sum _{k=m+1}^{n}a_{k}\right|<\epsilon }$

Here we require ${\displaystyle n>m\geq N}$ instead of ${\displaystyle n,m\geq N}$, because above we have only considered cases with ${\displaystyle n>m}$.

### Beautification of the formula

To make the above formula more beautiful, we set ${\displaystyle {\tilde {m}}=m+1}$. Thus ${\displaystyle n>m}$ becomes ${\displaystyle n\geq m+1={\tilde {m}}}$. Also ${\displaystyle m\geq N}$ becomes the inequality ${\displaystyle {\tilde {m}}=m+1\geq N+1}$. We get:

${\displaystyle \forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n\geq {\tilde {m}}\geq N+1:\left|\sum _{k={\tilde {m}}}^{n}a_{k}\right|<\epsilon }$

If we now set ${\displaystyle {\tilde {N}}=N+1}$:

${\displaystyle \forall \epsilon >0\,\exists {\tilde {N}}\in \mathbb {N} \,\forall n\geq {\tilde {m}}\geq {\tilde {N}}:\left|\sum _{k={\tilde {m}}}^{n}a_{k}\right|<\epsilon }$

We rename ${\displaystyle {\tilde {m}}\to m}$ und ${\displaystyle {\tilde {N}}\to N}$ and obtain:

${\displaystyle \forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n\geq m\geq N:\left|\sum _{k=m}^{n}a_{k}\right|<\epsilon }$

The above formula is thus true, if the series converges. It is called Cauchy criterion of series.

### Proof of the converse

We have shown that a convergent series satisfies the Cauchy criterion. Conversely a series that satisfies the Cauchy criterion is convergent, i.e. if ${\displaystyle \left|\sum _{k=m}^{n}a_{k}\right|}$ becomes arbitrary small. So let us assume that

${\displaystyle \forall \epsilon >0\,\exists {\tilde {N}}\in \mathbb {N} \,\forall n\geq {\tilde {m}}\geq {\tilde {N}}:\left|\sum _{k={\tilde {m}}}^{n}a_{k}\right|<\epsilon }$

is true, and convince ourself that the series must be convergent. In the above derivation we have seen that ${\displaystyle \left|\sum _{k=m}^{n}a_{k}\right|}$ is the distance ${\displaystyle |S_{n}-S_{m}|}$ for ${\displaystyle n>m}$ (after we renamed the variables). From the Cauchy criterion for series we can follow the Cauchy condition for the sequence of the partial sums with ${\displaystyle n>m}$. But we lack the Cauchy condition for the case ${\displaystyle m\geq n}$. Here we need to prove once more that ${\displaystyle |S_{n}-S_{m}|}$ becomes smaller than any ${\displaystyle \epsilon }$. We have

{\displaystyle {\begin{aligned}|S_{n}-S_{m}|&=|S_{m}-S_{n}|\\&=\left|\sum _{k=1}^{m}a_{k}-\sum _{l=1}^{n}a_{l}\right|\\[1em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{as above}}\right.}\\[1em]&=\left|\sum _{k=n+1}^{m}a_{k}\right|\end{aligned}}}

Again we have the absolute value of a sum of subsequent summands. We know from the Cauchy criterion for series that this value will become arbitrary small with increasing starting index and in particular it will be smaller than any ${\displaystyle \epsilon }$ if we choose a large enough starting index. So from the Cauchy criterion for series we have followed the Cauchy condition for the sequence of the partial sums ${\displaystyle (S_{n})_{n\in \mathbb {N} }}$, which by definition means convergence of the series.

## Definition of Cauchy Criterion

Definition (Cauchy criterion for series)

A series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ satisfies the Cauchy criterion for series, if the following is true:

${\displaystyle \forall \epsilon >0\,\exists N\in \mathbb {N} \,\forall n\geq m\geq N:\left|\sum _{k=m}^{n}a_{k}\right|<\epsilon }$

With annotations:

${\displaystyle \underbrace {{\underset {}{}}\forall \epsilon >0} _{{\text{For every distance }}\epsilon >0}\ \underbrace {{\underset {}{}}\exists N\in \mathbb {N} } _{{\text{ there exists a minimum index }}N}\ \underbrace {{\underset {}{}}\forall n\geq m\geq N:} _{{\text{so that for all indices }}n\geq m\geq N}\ \underbrace {{\underset {}{}}\left|\sum _{k=m}^{n}a_{k}\right|<\epsilon } _{{\text{ the absolute value }}\sum _{k=m}^{n}a_{k}{\text{ is smaller than }}\epsilon }}$

In the derivation we have also shown the following theorem:

Theorem (Cauchy criterion is equivalent to convergence)

A convergent series satisfies the Cauchy criterion, and conversely every real-valued series, that satisfies the Cauchy criterion, has a real values limit.

Question: Why do some textbooks define the Cauchy criterion with ${\displaystyle n\geq m>N}$ instead of ${\displaystyle n\geq m\geq N}$?

The two definitions are not in contradiction with each other: ${\displaystyle n\geq m>N}$ is equivalent to ${\displaystyle n\geq m\geq N+1}$. After renaming ${\displaystyle M=N+1}$ we obtain our known definition with ${\displaystyle n\geq m\geq M}$. The only difference between the definitions is whether we allow ${\displaystyle m=N}$ bzw. ${\displaystyle n=N}$ or not, the statement about convergence remains the same.

In practice the Cauchy criterion is not often used to determine the convergence of a given series. There are better convergence criteria to work with. But you will see the Cauchy criterion often used in proofs. For instance we can show that the Term test is correct, by using the Cauchy criterion. We can further show that any absolutely convergent series is also convergent in the normal way.

In the derivation we have also seen that the Cauchy criterion for series is analogue to the Cauchy criterion for sequences, but we apply it to the sequence of partial sums.

## Conclusion: Modifying finitely many summands has no influence on convergence

The Cauchy criterion for series tells us that changing the value of finitely many summands of the series will not affect the convergence behaviour. Take a series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$, and change finitely many summands. Let ${\displaystyle a_{N}}$ be the last summand that was changed. For all later ${\displaystyle n\geq m\geq N+1}$ the absolute value ${\displaystyle \left|\sum _{k=m}^{n}a_{k}\right|}$ will not change. So if the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ satisfies the Cauchy criterion, the same is true for the modified series and vice versa. Since the conformance to Cauchy criterion is equivalent to convergence, this means that the convergence behaviour of a series will not change if you modify a finite amount of summands (but the value of the series could change).

## Structure of proofs

### Convergence proof

The definition of the Cauchy criterion for a series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ is:

${\displaystyle {\color {Red}\forall \epsilon >0}\ {\color {RedOrange}\exists N\in \mathbb {N} }\ {\color {OliveGreen}\forall n\geq m\geq N}:{\color {Blue}\left|\sum _{k=m}^{n}a_{k}\right|<\epsilon }}$

From this definition we can derive the structure of a convergence proof:

${\displaystyle {\begin{array}{l}{\color {Red}\underbrace {{\underset {}{}}{\text{Let }}\epsilon >0{\text{ be arbitrary.}}} _{\forall \epsilon >0}}\ {\color {RedOrange}\underbrace {{\underset {}{}}{\text{Choose }}N=\ldots {\text{ There exists }}N{\text{, because}}\ldots } _{\exists N\in \mathbb {N} }}\\{\color {OliveGreen}\underbrace {{\underset {}{}}{\text{Let }}n\geq m\geq N{\text{ be arbitrary.}}} _{\forall n\geq m\geq N}}\ {\color {Blue}\underbrace {{\underset {}{}}{\text{We have: }}\left|\sum _{k=m}^{n}a_{k}\right|<\ldots <\epsilon } _{{}\left|\sum _{k=m}^{n}a_{k}\right|<\epsilon }}\end{array}}}$

When writing a convergence proof, you can use the above structure for orientation.

### Divergence proof

For divergence we can also have a proof structure using the Cauchy criterion. The formal definition:

${\displaystyle {\color {Red}\exists \epsilon >0}\ {\color {RedOrange}\forall N\in \mathbb {N} }\ {\color {OliveGreen}\exists n\geq m\geq N}:{\color {Blue}\left|\sum _{k=m}^{n}a_{k}\right|\geq \epsilon }}$

The structure for the proof is as follows:

${\displaystyle {\begin{array}{l}{\color {Red}\underbrace {{\underset {}{}}{\text{Choose }}\epsilon =\ldots } _{\exists \epsilon >0}}\ {\color {RedOrange}\underbrace {{\underset {}{}}{\text{Let }}N\in \mathbb {N} {\text{ be arbitrary.}}} _{\forall N\in \mathbb {N} }}\\{\color {OliveGreen}\underbrace {{\underset {}{}}{\text{Choose }}n=\ldots {\text{ and }}m=\ldots } _{\exists n\geq m\geq N}}\\{\color {Black}{\text{It follows that:}}}\\[0.5em]\quad \quad {\color {Blue}{\text{Proof for }}\left|\sum _{k=m}^{n}a_{k}\right|\geq \epsilon }\end{array}}}$

Hint

In proofs like these sometimes certain claims are trivial. Then you don't have to proof them separately. For example in some divergence proofs you can set ${\displaystyle \epsilon =1}$. You don't have to prove that ${\displaystyle \epsilon }$ exists, because it is clear that the number ${\displaystyle 1}$ exists.

## Finding a proof

The search for a proof will vary depending on how the final proof is structured. This is also true for proofs that make use of the Cauchy criterion. In this section we are going to explain possible ways of finding proofs with the Cauchy criterion.

### Convergence proof

The core of a convergence proof that uses the Cauchy criterion is the inequality chain ${\displaystyle \left|\sum _{k=m}^{n}a_{k}\right|<\ldots <\epsilon }$, i.e. no matter how small our ${\displaystyle \epsilon >0}$ is, we need to find a sufficiently large ${\displaystyle N\in \mathbb {N} }$, so that ${\displaystyle \left|\sum _{k=m}^{n}a_{k}\right|<\epsilon }$ for ${\displaystyle m\geq n\geq N}$. To find this inequaltiy chain we often estimate an upper bound for ${\displaystyle \left|\sum _{k=m}^{n}a_{k}\right|}$. We will have an inequality chain of the following form:

${\displaystyle \left|\sum _{k=m}^{n}a_{k}\right|\leq T_{1}(n,m)\leq T_{2}(n,m)\leq \ldots \leq T_{k}(n,m)}$

The terms ${\displaystyle T_{k}(n,m)}$ are dependent on ${\displaystyle m}$ and ${\displaystyle n}$. The goal is to reduce or simplfy these terms using clever estimates or rearrangement. But pay attention not to estimate too generously. All of the terms ${\displaystyle T_{k}(n,m)}$ has to be smaller than a chosen ${\displaystyle \epsilon >0}$, if ${\displaystyle m}$ and ${\displaystyle n}$ are sufficently large. For our estimates we can define arbitrary conditions ${\displaystyle m\geq N(\epsilon )}$ or ${\displaystyle n\geq N(\epsilon )}$, if this helps us. But unfortunately there are no general rules for those estimates. Sometimes you have to employ clever manipulations or (computational) tricks.

After we reduced the terms in the inequality chain enough, we should find that the last term ${\displaystyle T_{k}(n,m)}$ is smaller than ${\displaystyle \epsilon }$. We thus look at the inequality ${\displaystyle T_{k}(n,m)<\epsilon }$. Through equivalent reformulation of our condition we find ${\displaystyle m}$ and ${\displaystyle n}$, so that the term ${\displaystyle T_{k}(n,m)}$ is guaranteed to be smaller than ${\displaystyle \epsilon }$.

As the last step we have to choose our ${\displaystyle N}$. We do that by considering all conditions we found for ${\displaystyle m}$ and ${\displaystyle n}$. We first can restate our condition ${\displaystyle m\geq N(\epsilon )}$ as ${\displaystyle n\geq N(\epsilon )}$. Because of ${\displaystyle m\geq n}$ it follows from ${\displaystyle n\geq N(\epsilon )}$ that also ${\displaystyle m\geq N(\epsilon )}$. If you have the conditions ${\displaystyle n\geq N_{1}(\epsilon )}$, ${\displaystyle n\geq N_{2}(\epsilon )}$ up to ${\displaystyle n\geq N_{k}(\epsilon )}$, then we can set ${\displaystyle N=\max\{N_{1}(\epsilon ),N_{2}(\epsilon ),\ldots N_{k}(\epsilon )\}}$ in our final proof. From ${\displaystyle n\geq N=\max\{N_{1}(\epsilon ),\ldots N_{k}(\epsilon )\}}$ it follows that ${\displaystyle n}$ is greater than any ${\displaystyle N_{i}(\epsilon )}$. For example, imagine that for your inequality chain you need the following conditions:

• ${\displaystyle n\geq {\frac {1}{\epsilon }}}$
• ${\displaystyle n\geq \log(1+\epsilon )}$
• ${\displaystyle n\geq 42}$

Then you can simply set ${\displaystyle N=\max \left\{{\tfrac {1}{\epsilon }},\log(1+\epsilon ),42\right\}}$ in your proof.

### Divergence proof

To proof the divergence of a series through the Cauchy criterion we need to find a ${\displaystyle \epsilon }$, that satisfies the inequality chain ${\displaystyle \left|\sum _{k=m}^{n}a_{k}\right|\geq \ldots \geq \epsilon }$. In contrast to the convergence case our inequality must hold for all ${\displaystyle n,m\geq N}$ and for all ${\displaystyle N\in \mathbb {N} }$. For that we expand the summands of ${\displaystyle \left|\sum _{k=m}^{n}a_{k}\right|}$ and try to estimate a lower bound. We can make use of the free choice of ${\displaystyle n}$ which in relation to ${\displaystyle m}$ can be arbitrary large (only condition being ${\displaystyle n\geq m}$). In that way we can make ${\displaystyle \left|\sum _{k=m}^{n}a_{k}\right|}$ so big, that no watter what ${\displaystyle N\in \mathbb {N} }$ we choose the value of ${\displaystyle \left|\sum _{k=m}^{n}a_{k}\right|}$ will be bigger than any fixed positive real number. Set ${\displaystyle \epsilon }$ as the value of that number.

Again there are no general rules for the estimates and the choice ${\displaystyle n}$. What works oftentimes though, is to set all summands to the smallest value, and set ${\displaystyle n=tm}$ with respective ${\displaystyle t\in \mathbb {N} }$.

## Exercises

### Example for convergence

Exercise

Show that the series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}}$ is convergent using the Cauchy criterion[2].

How to get to the proof?

Here we employ two tricks. First, note that ${\displaystyle {\tfrac {1}{k}}<{\tfrac {1}{k-1}}}$. Also we use the equation ${\displaystyle {\tfrac {1}{k(k-1)}}={\tfrac {1}{k-1}}-{\tfrac {1}{k}}}$. So we obtain

${\displaystyle \sum _{k=m}^{n}{\frac {1}{k^{2}}}<\sum _{k=m}^{n}{\frac {1}{k(k-1)}}=\sum _{k=m}^{n}\left({\frac {1}{k-1}}-{\frac {1}{k}}\right)}$

The sum ${\displaystyle \sum _{k=m}^{n}\left({\tfrac {1}{k-1}}-{\tfrac {1}{k}}\right)}$ is a telescoping sum, thus we can resolve the sum:

{\displaystyle {\begin{aligned}\sum _{k=m}^{n}\left({\frac {1}{k-1}}-{\frac {1}{k}}\right)&=\left({\frac {1}{m-1}}-{\frac {1}{m}}\right)+\left({\frac {1}{m}}-{\frac {1}{m+1}}\right)+\ldots \left({\frac {1}{n-1}}-{\frac {1}{n}}\right)\\[1em]&={\frac {1}{m-1}}+\left(-{\frac {1}{m}}+{\frac {1}{m}}\right)+\left(-{\frac {1}{m+1}}+{\frac {1}{m+1}}\right)+\ldots -{\frac {1}{n}}\\[1em]&={\frac {1}{m-1}}-{\frac {1}{n}}\\[1em]&<{\frac {1}{m-1}}\end{aligned}}}

Overall we can thus prove ${\displaystyle \sum _{k=m}^{n}{\tfrac {1}{k^{2}}}<{\tfrac {1}{m-1}}}$. The right term ${\displaystyle {\tfrac {1}{m-1}}}$ goes towards zero, if ${\displaystyle m\to \infty }$. That way we can show the Cauchy criterion.

Proof

Let ${\displaystyle \epsilon >0}$ be arbitrary. Choose ${\displaystyle N\in \mathbb {N} }$, so that ${\displaystyle {\tfrac {1}{N-1}}<\epsilon }$. This ${\displaystyle N}$ exists, because ${\displaystyle \lim _{N\to \infty }{\tfrac {1}{N-1}}=0}$. Then, let ${\displaystyle n\geq m\geq N}$ be arbitrary. We have:

{\displaystyle {\begin{aligned}\sum _{k=m}^{n}{\frac {1}{k^{2}}}&<\sum _{k=m}^{n}{\frac {1}{k(k-1)}}\\[0.5em]&=\sum _{k=m}^{n}\left({\frac {1}{k-1}}-{\frac {1}{k}}\right)\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ {\text{Teleskop-Summe}}\right.}\\[0.5em]&={\frac {1}{m-1}}-{\frac {1}{n}}\\[0.5em]&<{\frac {1}{m-1}}\\[0.5em]&\quad {\color {OliveGreen}\left\downarrow \ m\geq N\right.}\\[0.5em]&\leq {\frac {1}{N-1}}\\[0.5em]&<\epsilon \end{aligned}}}

Thus the series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}}$ satisfies the Cauchy criterion for series and is convergent.

### Example for divergence

Exercise

Show that the harmonic series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k}}}$ is divergent using the Cauchy criterion.[3].

How to get to the proof?

Here again we use a trick. We first observe that

{\displaystyle {\begin{aligned}{\frac {1}{n+1}}+{\frac {1}{n+2}}+\ldots +{\frac {1}{2n}}&\geq {\frac {1}{2n}}+{\frac {1}{2n}}+\ldots +{\frac {1}{2n}}\\[0.5em]&=n\cdot {\frac {1}{2n}}\\[0.5em]&={\frac {1}{2}}\end{aligned}}}

Thus we see

${\displaystyle \left|\sum _{k=n+1}^{2n}{\frac {1}{k}}\right|={\frac {1}{n+1}}+\ldots +{\frac {1}{2n}}\geq {\frac {1}{2}}}$

This shows that ${\displaystyle \left|\sum _{k=n+1}^{2n}{\frac {1}{k}}\right|}$ cannot become arbitrary small, so the Cauchy criterion cannot be satisfied.

Proof

Let ${\displaystyle 0<\epsilon \leq {\tfrac {1}{2}}}$ arbitrary (e.g. ${\displaystyle \epsilon ={\tfrac {1}{4}}}$). Let ${\displaystyle N\in \mathbb {N} }$ be given. For ${\displaystyle n\geq N}$ we now have:

{\displaystyle {\begin{aligned}\left|\sum _{k=n+1}^{2n}{\frac {1}{k}}\right|&={\frac {1}{n+1}}+{\frac {1}{n+2}}+\ldots +{\frac {1}{2n}}\\[0.5em]&\geq {\frac {1}{2n}}+{\frac {1}{2n}}+\ldots +{\frac {1}{2n}}\\[0.5em]&=n\cdot {\frac {1}{2n}}\\[0.5em]&={\frac {1}{2}}\\&\geq \epsilon \end{aligned}}}

There cannot be any ${\displaystyle N\in \mathbb {N} }$, so that ${\displaystyle \left|\sum _{k=n+1}^{2n}{\frac {1}{k}}\right|}$ is smaller than ${\displaystyle \epsilon }$ for ${\displaystyle n\geq N}$. This shows that the series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k}}}$ cannot satisfy the Cauchy criterion for series. So this means that this series is divergent.

For another example of a convergence case, take a look at the alternating harmonic series in the respective Exercise.