# How to prove convergence and divergence – Serlo

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In this chapter, we will explain how convergence and divergence of a sequence can be proven. Usually, this job splits into two steps: At first, one tries some brainstorming (with a pencil on a piece of paper), trying to find a way to prove convergence or divergence. Then, if one has a solution, one tries to write it down in a short and elegant way. What you can read in most math books is the result of the second step. The aim of step 2 ist to conserve your thoughts for further people (or a later version of yourself), such that the reader can understand the proof investing as little time / effort as possible. However, the thoughts which a mathematician has when trying to find a proof (in step 1) are often quite different from what is written down in step 2. The problems given in this chapter will illustrate both steps and the difference between how to get to a proof and how to write it down.

Be aware that for proving convergence or divergence, there is no "cooking recipe", which will always lead you to a working proof! There is rather a collection of tools you can always carry around with you (like a Swiss pocket knife) and which you can use for mathematical problem solving. In this chapter, we would like to provide you with a specific collection of such tools. Not every problem you encounter in an exercise class is directly solvable with one tool and sometimes, you need to creatively combine different tools and techniques in order to crack open a problem. Those exercises are designed for purpose: They are intended to train your skills in solving abstract problems by combining strategies you know in creative and uncommon ways. These problem solving skills turn out to be useful in a broad variety of jobs and even for your personal life. This is the reason, why math courses are part of the syllabus in school and a huge variety of university degree programmes! (Torturing you with technical formulas is only secondary ).

## Proving convergence

An example how to prove convergence of a sequence (video in German)

### General structure of proof

Before turning to examples, we take a closer look at the structure of the proof. This way, we know what the final proof should look like. There is a mathematical definition for convergence of a sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ to a limit ${\displaystyle a}$, which looks as follows:

${\displaystyle {\color {Red}\forall \epsilon >0}\ {\color {RedOrange}\exists N\in \mathbb {N} }\ {\color {OliveGreen}\forall n\geq N}\ {\color {Blue}|a_{n}-a|<\epsilon }}$

I.e. after passing an index number ${\displaystyle N}$, all elements with index number ${\displaystyle n}$ coming afterwards will be closer to the limit than a small amount ${\displaystyle \epsilon }$. The proof that this statement holds looks as follows:

${\displaystyle {\begin{array}{l}{\color {Red}\underbrace {{\underset {}{}}{\text{Let }}\epsilon >0{\text{ be arbitrary.}}} _{\forall \epsilon >0}}\ {\color {RedOrange}\underbrace {{\underset {}{}}{\text{Choose some }}N=\ldots {\text{ Such an }}N{\text{ exists, because}}\ldots } _{\exists N\in \mathbb {N} }}\\{\color {OliveGreen}\underbrace {{\underset {}{}}{\text{Let then be }}n\geq N} _{\forall n\geq N}}\ {\color {Blue}{\text{Fore those numbers, there is: }}|a_{n}-a|<\ldots <\epsilon }\end{array}}}$

If you explicitly write down, what the natural number ${\displaystyle N}$ looks like, you do not need to explicitly show that ${\displaystyle N}$ exists, because writing down an explicit expression is already a proof.

### An example problem

"Does the sequence ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ with ${\displaystyle a_{n}={\tfrac {n}{n+1}}}$ converge? If yes, what is the limit? Prove all your claims about convergence/ divergence."

The solution involves the following steps:

1. Investigate the sequence for high ${\displaystyle n}$, in order to find a limit ${\displaystyle a}$
2. Make up a proof that ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ converges to ${\displaystyle a}$ with a pencil on a piece of paper
3. Write down the proof in a short way, structured as above

### Finding a limit

How does the sequence ${\displaystyle a_{n}={\tfrac {n}{n+1}}}$ behave for high ${\displaystyle n}$? Will it converge? In order to get more confident with the sequence, there are some actions you could try:

• Calculate the first couple of elements: Compute the first couple of elements and what they explicitly look like. It is also useful to draw them in a diagram. Do they increase or decrease faster and faster? Are they jumping between two values? Or do they seem to approach a certain value?
• Compute some elements with huge indices: What is ${\displaystyle a_{1000}}$? How about ${\displaystyle a_{1000000}}$? A calculator or PC can be useful to find that out. Or you can make rough estimates what those ${\displaystyle a_{n}}$ are. Do they come close to a specific real number ${\displaystyle a}$? Then this may be your limit.
• Make assertions based in intuition: The more problems you solve, the more you get an intuition what a limit might be. Polynomials ${\displaystyle n^{k}}$ always run to infinity, as ${\displaystyle n\to \infty }$. Exponentials ${\displaystyle e^{n}}$ run away even faster. Sequences like ${\displaystyle {\tfrac {1}{n^{k}}}}$ and ${\displaystyle e^{-n}}$ tend to 0. For fractions ${\displaystyle {\tfrac {n^{a}}{n^{b}}}}$, the higher power wins and exponential functions win against everything (if you feel confused now: we will later explain in detail what these intuitions mean).

So let us start to compute the first (let's say 10) elements of the sequence ${\displaystyle a_{n}={\tfrac {n}{n+1}}}$ :

${\displaystyle n}$ ${\displaystyle {\tfrac {n}{n+1}}}$
1 0,5
2 0,666…
3 0,75
4 0,8
5 0,833…
6 0,857…
7 0,875
8 0,888…
9 0,9
10 0,909…

We could also plot them in a diagram:

The elements seem to be monotonously increasing. The increase is getting smaller and smaller. Perhaps, it converges. But what can be a possible limit? Maybe, we compute some elements with high index numbers to see it:

${\displaystyle a_{1000}=0,99900099900\ldots }$

or even higher:

${\displaystyle a_{1000000}=0,99999900000099\ldots }$

These numbers are suspiciously close to ${\displaystyle 1}$ . So we may assert that ${\displaystyle 1}$ is the limit of the sequence. This intuitively makes sense in the following way: For high ${\displaystyle n}$ , there is ${\displaystyle n+1\approx n}$ (${\displaystyle 1000001}$ is almost the same as ${\displaystyle 1000000}$). So for the quotient, we expect

${\displaystyle {\tfrac {n}{n+1}}\approx {\tfrac {n}{n}}=1}$

following these considerations, we make the assertion that ${\displaystyle 1}$ is the limit of the sequence ${\displaystyle a_{n}={\tfrac {n}{n+1}}}$.

Warning

The argumentation above is not a full mathematical proof. We only assert that 1 is the limit, but we don't know it yet. Writing down some assertions is a helpful technique but does not score you any points in an exam. You always need a proof in order to consider a problem as solved.

### Finding the proof steps

The proof builds around establishing an inequality of the form ${\displaystyle |a_{n}-a|<\ldots <\epsilon }$. It is best to start with ${\displaystyle |a_{n}-a|}$ and try to find greater and greater terms, until one of them is ${\displaystyle \epsilon }$ . Throughout these estimates, we can make any assumptions on ${\displaystyle n}$ of the form ${\displaystyle n\geq N}$ with ${\displaystyle N}$ being a natural number only depending on ${\displaystyle \epsilon }$ and ${\displaystyle a}$ (note ${\displaystyle N}$ must not depend on ${\displaystyle a_{n}}$ !). If ${\displaystyle N}$ is too small, we just choose a bigger ${\displaystyle N}$.

Question: Why is ${\displaystyle N}$ not allowed to depend on ${\displaystyle a_{n}}$?

In the proof structure above, we see that ${\displaystyle n}$ is defined such that ${\displaystyle n\geq N}$ . So it depends on ${\displaystyle N}$, which values for ${\displaystyle n}$ are allowed. If one makes ${\displaystyle N}$ depending on ${\displaystyle n}$ , we would have some circle of dependences, which may or may not be resolved. For instance, if we have ${\displaystyle a_{n}=2n}$ and choose ${\displaystyle N, then there is ${\displaystyle N<2n}$, which does trivially hold for all ${\displaystyle n>N}$. However, a condition like ${\displaystyle N>a_{n}/2}$ amounts to ${\displaystyle N>n}$, which can never be fulfilled for any ${\displaystyle n>N}$. So we might run into the trap of imposing a condition on ${\displaystyle n}$, which can never be fulfilled.

In the proof structure above, we see that the ${\displaystyle N}$ is defined before considering an ${\displaystyle n}$ or ${\displaystyle a_{n}}$ . So ${\displaystyle N}$ should not depend on ${\displaystyle n}$.

We can also recall the definition of convergence in order to see the dependence issue:

${\displaystyle {\color {Blue}\exists a\in \mathbb {R} \,\forall \epsilon >0}\ {\color {Orange}\exists N\in \mathbb {N} }\ {\color {OliveGreen}\forall n\geq N}\ |a_{n}-a|<\epsilon }$

${\displaystyle {\color {Orange}N}}$ may only depend on what has been fixed before, i.e. what is on the left side of it: ${\displaystyle {\color {Blue}a}}$ and ${\displaystyle {\color {Blue}\epsilon }}$. By contrast, the variable ${\displaystyle {\color {OliveGreen}n}}$ is being introduced after ${\displaystyle {\color {Orange}N}}$ and stands on the right of it. Therefore, ${\displaystyle {\color {Orange}N}}$ may neither depend on ${\displaystyle {\color {OliveGreen}n}}$ nor on ${\displaystyle a_{\color {OliveGreen}n}}$ .

Another often successful technique is to take ${\displaystyle |a_{n}-a|<\epsilon }$ and get ${\displaystyle n}$ standing alone on one side. That means, we use some equivalent reformulations of the term in order to get it in the form ${\displaystyle n\geq \ldots }$ with ${\displaystyle \dots }$ some term depending on ${\displaystyle \epsilon }$. If we choose ${\displaystyle N}$ such that ${\displaystyle N\geq \ldots }$, then for any ${\displaystyle n>N}$, there is also ${\displaystyle n\geq \ldots }$. ${\displaystyle n\geq \ldots }$ is just equivalent to ${\displaystyle |a_{n}-a|<\epsilon }$ (if we only use equivalent reformulations, of course). So we have found a suitable ${\displaystyle N}$, such that ${\displaystyle |a_{n}-a|<\epsilon }$ for all ${\displaystyle n>N}$. The following chapters will provide some examples how these techniques are applied.

Sometimes, we encounter several conditions on ${\displaystyle n}$, such that ${\displaystyle |a_{n}-a|<\epsilon }$ can be made sure to hold, for instance ${\displaystyle n\geq N_{1}(a,\epsilon )}$, ${\displaystyle n\geq N_{2}(a,\epsilon )}$,…,${\displaystyle n\geq N_{m}(a,\epsilon )}$. In those cases, we just take the maximum ${\displaystyle N=\max\{N_{1}(a,\epsilon ),N_{2}(a,\epsilon ),\ldots ,N_{m}(a,\epsilon )\}}$. We can imagine ${\displaystyle N_{1},N_{2},...,N_{m}}$ as thresholds which ${\displaystyle n}$ has to cross in order that ${\displaystyle a_{n}}$ approaches ${\displaystyle a}$ up to ${\displaystyle \epsilon }$. After ${\displaystyle n}$ has passed the highest of the thresholds ${\displaystyle N}$, all other thresholds will have been passed, too and all conditions for ${\displaystyle |a_{n}-a|<\epsilon }$ are met. Usually, one makes up the conditions with ${\displaystyle N_{1},N_{2},...,N_{m}}$ in step 1 on a piece of paper, whereas when writing down the proof in step 2, one only defines ${\displaystyle N}$ (without giving the derivation).

Now let us return to the example. It is very useful to equivalently reformulate ${\displaystyle \left|1-{\tfrac {n}{n+1}}\right|}$:

{\displaystyle {\begin{aligned}\left|1-{\frac {n}{n+1}}\right|&=\left|{\frac {n+1}{n+1}}-{\frac {n}{n+1}}\right|=\left|{\frac {n+1-n}{n+1}}\right|\\[0.5em]&=\left|{\frac {1}{n+1}}\right|={\frac {1}{n+1}}\end{aligned}}}

We know that this term must get lower than any ${\displaystyle \epsilon >0}$ for sufficiently high ${\displaystyle n}$. This is sometimes also called Archimedean axiom: for all ${\displaystyle \epsilon >0}$ there exists an ${\displaystyle M\in \mathbb {N} }$ with ${\displaystyle {\tfrac {1}{M}}<\epsilon }$ . We can directly reach ${\displaystyle {\tfrac {1}{n+1}}<\epsilon }$ by choosing ${\displaystyle n+1\geq M}$ , since we instantly get ${\displaystyle {\tfrac {1}{n+1}}\leq {\tfrac {1}{M}}<\epsilon }$. Hence, it suffices if ${\displaystyle n}$ meets the following condition:

${\displaystyle {\begin{array}{rrl}&n+1&\geq M\\\Leftrightarrow \ &n&\geq M-1\end{array}}}$

So we found the desired bound with condition ${\displaystyle n\geq M-1}$. For writing down the proof, we choose ${\displaystyle N=M-1}$, with ${\displaystyle M}$ being the fixed number from the Archimedean axiom above.

### Writing down the proof

Now, we go over to step 2 and write down the proof. The final solution is quite short and concise:

Proof

Let ${\displaystyle \epsilon >0}$ be arbitrary. By the Archimedean axiom, there is an ${\displaystyle M\in \mathbb {N} }$ with ${\displaystyle {\tfrac {1}{M}}<\epsilon }$. We choose ${\displaystyle N=M-1}$. For all ${\displaystyle n\geq N}$ there is:

{\displaystyle {\begin{aligned}\left|1-{\frac {n}{n+1}}\right|&=\left|{\frac {n+1}{n+1}}-{\frac {n}{n+1}}\right|=\left|{\frac {n+1-n}{n+1}}\right|\\[0.5em]&=\left|{\frac {1}{n+1}}\right|={\frac {1}{n+1}}\leq {\frac {1}{M}}<\epsilon \end{aligned}}}

That's it already! If we compare the proof above (step 2) with the derivation (step 1), we notice that they look entirely different. The proof is enormously short and the ${\displaystyle M}$ and ${\displaystyle N}$ seem to appear out of nowhere! It ma appear to you that the mathematician who wrote the proof is some kind of superhuman genius who has found some magical way to always and instantly get the right ${\displaystyle M}$ and ${\displaystyle N}$. However, this is not the case. In the beginning, the author often had no idea how the proof works and performed a lot of trial and error, as well as lengthy considerations in step 1. He/ she just did not write them down, because they would just require additional space on the paper and additional time to read.

### Übungsaufgabe

Wir empfehlen euch, genau wie eben beschrieben, die folgende Aufgabe zu versuchen.

Exercise (Konvergenz einer Folge)

Beweise, dass die Folge ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ mit ${\displaystyle a_{n}={\tfrac {1-2n}{5+3n}}}$ konvergiert. Wie lautet ihr Grenzwert?

How to get to the proof? (Konvergenz einer Folge)

Wir gegehen genau wie oben beschrieben vor. Zunächst benötigen wir einen Grenzwert:

Lösungsschritt: Grenzwert finden

Wir können wie oben die ersten Folgenglieder ausrechnen, oder wir überlegen uns gleich folgendes: Für sehr große ${\displaystyle n}$ gilt für den Zähler der Folge ${\displaystyle 1-2n\approx -2n}$, und für den Nenner ${\displaystyle 5+3n\approx 3n}$. Insgesamt gilt daher

${\displaystyle a_{n}={\frac {1-2n}{5+3n}}\approx {\frac {-2n}{3n}}=-{\frac {2}{3}}}$

falls ${\displaystyle n}$ sehr groß ist. Unsere starke Vermutung ist somit, dass ${\displaystyle (a_{n})}$ gegen den Grenzwert ${\displaystyle -{\tfrac {2}{3}}}$ konvergiert.

Nun folgt die rechnerische Vorarbeit, um anschließend den Beweis sauber aufschreiben zu können:

Lösungsschritt: Nötige Beweisschritte finden

Laut der Definition der Konvergenz müssen wir zu jedem ${\displaystyle \epsilon >0}$ ein ${\displaystyle N\in \mathbb {N} }$ finden, so dass für alle ${\displaystyle n\geq N}$ gilt: ${\displaystyle |a_{n}-(-{\tfrac {2}{3}})|<\epsilon }$. Dazu vereinfachen wir den Ausdruck ${\displaystyle |a_{n}-(-{\tfrac {2}{3}})|}$ zunächst:

{\displaystyle {\begin{aligned}|a_{n}-(-{\tfrac {2}{3}})|&=\left|{\frac {1-2n}{5+3n}}+{\frac {2}{3}}\right|\\[0.3em]&=\left|{\frac {3(1-2n)}{3(5+3n)}}+{\frac {2(5+3n)}{3(5+3n)}}\right|\\[0.3em]&=\left|{\frac {3(1-2n)+2(5+3n)}{3(5+3n)}}\right|\\[0.3em]&=\left|{\frac {3-6n+10+6n}{15+9n}}\right|\\[0.3em]&=\left|{\frac {13}{15+9n}}\right|={\frac {13}{15+9n}}\end{aligned}}}

Nun formen wir die Ungleichung ${\displaystyle {\tfrac {13}{15+9n}}<\epsilon }$ um, zu einer Ungleichung der Form ${\displaystyle n>\ldots }$:

{\displaystyle {\begin{aligned}&{\frac {13}{15+9n}}<\epsilon \\[0.3em]\iff &13<\epsilon (15+9n)\\[0.3em]\iff &13<15\epsilon +9n\epsilon \\[0.3em]\iff &13-15\epsilon <9n\epsilon \\[0.3em]\iff &{\frac {13-15\epsilon }{9\epsilon }}{\frac {13-15\epsilon }{9\epsilon }}\end{aligned}}}

Damit haben wir eine passende Bedingung für ${\displaystyle n}$, und damit auch ${\displaystyle N}$ gefunden. Wählen wir nämlich ${\displaystyle N>{\tfrac {13-15\epsilon }{9\epsilon }}}$, was nach dem archimedischen Axiom möglich ist, so folgt aus dem eben hergeleiteten für alle ${\displaystyle n\geq N}$: ${\displaystyle |a_{n}-(-{\tfrac {2}{3}})|={\tfrac {13}{15+9n}}<\epsilon }$.

Damit sind wir mit unserer Vorarbeit fertig, und müssen den Beweis nur noch in „Mathematikerdeutsch“ formulieren.

Proof (Konvergenz einer Folge)

Sei ${\displaystyle \epsilon >0}$ beliebig. Nach dem archimedischen Axiom gibt es ein ${\displaystyle N\in \mathbb {N} }$ mit ${\displaystyle N>{\tfrac {13-15\epsilon }{9\epsilon }}}$. Sei ${\displaystyle n\geq N}$ beliebig. Dann ist

{\displaystyle {\begin{aligned}\left|{\frac {1-2n}{5+3n}}-\left(-{\frac {2}{3}}\right)\right|&=\left|{\frac {3(1-2n)}{3(5+3n)}}+{\frac {2(5+3n)}{3(5+3n)}}\right|=\left|{\frac {13}{15+9n}}\right|\\[0.5em]&={\frac {13}{15+9n}}<\epsilon \end{aligned}}}

## Beweise für Divergenz führen

### Allgemeine Beweisstruktur

Die Divergenz einer Folge tritt per Definition genau dann ein, wenn die Folge nicht konvergent ist. Die aussagenlogische Formulierung von Divergenz ist also genau die Negation der Konvergenz-Definition. Dafür tauschen wir alle Quantoren aus und ändern im Teil nach den Quantoren ${\displaystyle <}$ zu ${\displaystyle \geq }$. (Analog würden wir bei Negation ${\displaystyle >}$ zu ${\displaystyle \leq }$ und ${\displaystyle =}$ zu ${\displaystyle \neq }$ umändern.) Bei Divergenz der Folge ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ haben wir also folgende Aussage zu beweisen:

${\displaystyle {\color {Red}\forall a\in \mathbb {R} }\ {\color {RedOrange}\exists \epsilon >0}\ {\color {DarkOrchid}\forall N\in \mathbb {N} }\ {\color {OliveGreen}\exists n\geq N}{\color {Blue}|a_{n}-a|\geq \epsilon }}$

Die damit verbundene Beweisstruktur ist:

${\displaystyle {\begin{array}{l}{\color {Red}\underbrace {{\underset {}{}}{\text{Sei }}a\in \mathbb {R} {\text{ beliebig.}}} _{\forall a\in \mathbb {R} }}\ {\color {RedOrange}\underbrace {{\underset {}{}}{\text{Wähle }}\epsilon =\ldots {\text{ Die Zahl }}\epsilon {\text{ existiert, weil}}\ldots } _{\exists \epsilon >0}}\\{\color {DarkOrchid}\underbrace {{\underset {}{}}{\text{Sei }}N\in \mathbb {N} {\text{ beliebig.}}} _{\forall N\in \mathbb {N} }}\ {\color {OliveGreen}\underbrace {{\underset {}{}}{\text{Wähle }}n=\ldots {\text{ Es existiert }}n{\text{ mit }}n\geq N{\text{, weil}}\ldots } _{\exists n\geq N}}\\{\color {Blue}{\text{Es ist: }}|a_{n}-a|\geq \ldots \geq \epsilon }\end{array}}}$

Hier können Teile des Beweisschemas weggelassen werden, wenn sie offensichtlich sind. Jedoch muss die grundlegende Beweisstruktur erhalten bleiben.

### Beispielaufgabe

Die Folge ${\displaystyle (2^{n})_{n\in \mathbb {N} }}$

Schauen wir uns den Divergenzbeweis exemplarisch an folgender Aufgabe an:

„Divergiert die Folge ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ mit ${\displaystyle a_{n}=2^{n}}$? Beweise deine Behauptung.“

Auch hier können wir mit den obigen Techniken (erste Folgenglieder berechnen, große Folgenglieder ausrechnen usw.) eine Vermutung aufstellen, ob diese Folge divergiert. Wir sehen aber schnell, dass die Folge über alle Grenzen hinweg wächst und sich dabei keiner reellen Zahl annähert. Die Folge ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ sollte also divergieren. Jetzt versuchen wir, einen Beweis für diese Behauptung zu finden.

### Lösungsweg

Kern des späteren Beweises ist die zu zeigende Ungleichungskette

${\displaystyle |a_{n}-a|\geq \ldots \geq \epsilon }$

Starten wir also wieder mit dem Betrag ${\displaystyle |a_{n}-a|}$. Auf einem Schmierblatt versuchen wir diesen Ausdruck so lange zu vereinfachen und nach unten abzuschätzen, bis wir einen Term ${\displaystyle \epsilon >0}$ haben. ${\displaystyle a}$ ist dabei beliebig vorgegeben und wir können keinen Einfluss auf den Wert von ${\displaystyle a}$ nehmen. Schließlich müssen wir den Beweis für alle Zahlen ${\displaystyle a\in \mathbb {R} }$ führen.

Jedoch können wir ${\displaystyle \epsilon }$ und ${\displaystyle n}$ frei wählen. Es muss nur gesichert sein, dass ${\displaystyle \epsilon >0}$ und ${\displaystyle n\geq N}$ ist, wobei ${\displaystyle N}$ eine beliebige natürliche Zahl ist. Da ${\displaystyle \epsilon }$ nach ${\displaystyle a}$ im Beweis eingeführt wird, darf ${\displaystyle \epsilon }$ von ${\displaystyle a}$ abhängen (jedoch nicht von ${\displaystyle n}$). Die natürliche Zahl ${\displaystyle n}$ darf sowohl von ${\displaystyle a}$, als auch von ${\displaystyle \epsilon }$ abhängen. Wir können also während der Abschätzung nach unten beliebige Bedingungen an ${\displaystyle \epsilon }$ und ${\displaystyle n}$ sammeln. Diese Bedingungen werden zum Schluss ähnlich wie beim Konvergenzbeweis zusammengefasst.

Fangen wir also an mit ${\displaystyle |2^{n}-a|}$. Um den Term zu vereinfachen, können wir ${\displaystyle 2^{n}\geq a}$ fordern, weil wir dann den Betrag weglassen können. Dass für ein ${\displaystyle n\in \mathbb {N} }$ die Ungleichung ${\displaystyle 2^{n}\geq a}$ erfüllt ist, erhalten wir aus den Folgerungen der Bernoulli-Ungleichung. Eine davon besagt:

„Für jede Zahl ${\displaystyle p>1}$ und jede Zahl ${\displaystyle M\in \mathbb {R} }$ gibt es ein ${\displaystyle n\in \mathbb {N} }$, so dass ${\displaystyle p^{n}>M}$ ist.“

Wir müssen nur ${\displaystyle M=a}$ und ${\displaystyle p=2}$ setzen. So erhalten wir mit der Bedingung ${\displaystyle 2^{n}\geq a}$:

${\displaystyle |2^{n}-a|=2^{n}-a}$

Nun müssen wir ${\displaystyle 2^{n}-a\geq \epsilon }$ beweisen, also formen wir dies durch Äquivalenzumformungen um:

${\displaystyle 2^{n}-a\geq \epsilon \iff 2^{n}\geq a+\epsilon }$

So erhalten wir die neue Bedingung ${\displaystyle 2^{n}\geq a+\epsilon }$, womit wir die letzte Ungleichung beweisen können. Für ${\displaystyle \epsilon }$ haben wir noch keine Bedingungen und können damit diese Zahl frei wählen. Dass es tatsächlich für jedes ${\displaystyle \epsilon }$ ein ${\displaystyle n}$ gibt mit ${\displaystyle 2^{n}\geq a+\epsilon }$ , liegt daran, dass wir die Folgerung aus der Bernoulli-Ungleichung auch mit ${\displaystyle M=a+\epsilon }$ benutzen können. Wir müssen nur aufpassen, dass ${\displaystyle \epsilon >0}$ ist. So wählen wir einfach ${\displaystyle \epsilon =1}$. Für ${\displaystyle n}$ haben wir die beiden Bedingungen ${\displaystyle 2^{n}\geq a}$ und ${\displaystyle 2^{n}\geq a+\epsilon =a+1}$. Also wählen wir ${\displaystyle 2^{n}\geq \max\{a,\,a+1\}=a+1}$, um beide Bedingungen zusammenzufassen.

### Beweis aufschreiben

Nun haben wir alle notwendigen Schritte, um den Beweis zu führen:

Proof

Sei ${\displaystyle a\in \mathbb {R} }$ beliebig. Wähle ${\displaystyle \epsilon =1}$. Sei ${\displaystyle N\in \mathbb {N} }$ beliebig. Wähle ${\displaystyle n\geq N}$ so, dass ${\displaystyle 2^{n}\geq a+1}$ ist. Dies ist aufgrund der Folgerungen aus der Bernoulli-Ungleichung möglich. Es ist nun

{\displaystyle {\begin{aligned}|2^{n}-a|&=2^{n}-a\geq (a+1)-a=1=\epsilon \end{aligned}}}

## Weitere Beweismethoden für Konvergenz und Divergenz

In den obigen beiden Abschnitten haben wir die Konvergenz beziehungsweise die Divergenz einer Folge direkt über die Epsilon-Definition des Grenzwerts geführt. In den folgenden Kapiteln wirst du auch folgende weitere Möglichkeiten kennen lernen, mit denen du Beweise zur Konvergenz und Divergenz führen kannst:

• Jede unbeschränkte und monotone Folge divergiert. Beispiel: Die Folge ${\displaystyle a_{n}=n}$ divergiert, weil sie unbeschränkt ist.
• Jede beschränkte und monotone Folge konvergiert. Beispiel: Die Folge ${\displaystyle a_{n}=1-{\tfrac {1}{2^{n}}}}$ ist nach unten durch ${\displaystyle {\tfrac {1}{2}}}$ und nach oben durch ${\displaystyle 1}$ beschränkt. Außerdem ist die Folge monoton steigend. Deswegen konvergiert sie.
• Sei ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ eine Folge. Wenn es für alle ${\displaystyle \epsilon >0}$ ein ${\displaystyle N\in \mathbb {N} }$ gibt, so dass ${\displaystyle |a_{n}-a_{m}|<\epsilon }$ für alle ${\displaystyle n,m\geq N}$ ist, dann konvergiert die Folge ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$. Dieses Kriterium wird Cauchy-Kriterium genannt, und ich werde es dir später genauer vorstellen. Es wird hauptsächlich in allgemeineren Beweisen verwendet und weniger dazu, die Konvergenz einer speziellen Folge zu zeigen.
• Mit Hilfe der Grenzwertsätze und des Sandwichsatzes kannst du auch den Grenzwert von Folgen berechnen, ohne die Epsilon-Definition des Grenzwerts verwenden zu müssen.