Convergence of Real Sequences

Convergence and Real Limits

Definition: Convergence of Real Sequences

Let $\{a_n\}$ be a real sequence.

We say that $\{a_n\}$ converges to $L\in \mathbb{R}$ iff for each $\epsilon >0$, there exists some integer $N$ such that

$$|a_n-L|<\epsilon \forall n\ge N$$

The number $L$, if it exists, is called the limit of $\{a_n\}$ as $n$ approaches infinity.

NOTATION

The most common notation is

$$\underset{n\to \infty }{\lim }{a}_n=L$$

In text, one also writes ”$a_n\to L$ as $n\to \infty$” or just "$a_n\to L$". Sometimes, one might also encounter $a_n\underset{n\to \infty }{⟶}L$ and $a_n\stackrel{n\to \infty }{⟶}L$.

Theorem: Uniqueness of the Limit

The limit of a real sequence, if it exists, is unique - if $\{a_n\}$ converges to both $L$ and $M$, then $L=M$.

PROOF

Pick some arbitrary $\epsilon >0$.

Since $\{a_n\}\to L$, by definition, there exists some integer $N_L$ such that

$$|a_n-L|<\epsilon \forall n\ge N_L.$$

Similarly, since $\{a_n\}\to M$, there exists some integer $N_M$ such that

$$|a_n-M|<\epsilon \forall n\ge N_M.$$

Now, let $N=\max \{N_L,N_M\}$. For all $n\ge N$, both of the aforementioned inequalities hold. Therefore, for all $n\ge N$, we have

$$|L-M|=|L-a_n+a_n-M|\le |L-a_n|+|a_n-M|<\epsilon +\epsilon =2\epsilon .$$

Therefore,

$$\frac{1}{2}|L-M|<\epsilon$$

So far, the argument does not actually depend on the particular choice of $\epsilon$ and is thus true for every $\epsilon >0$. This means that $\frac{1}{2}|L-M|$ is smaller that every positive real number. This is only possible if $\frac{1}{2}|L-M|$ is zero which is in turn only possible if $|L-M|=0$. Therefore, $L=M$.

Characterizations

Theorem: Approaching Zero

A real sequence $\{a_n\}$ converges to $L\in \mathbb{R}$ if and only if

$$\underset{n\to \infty }{\lim }|a_n-L|=0$$

PROOF

TODO

Theorem: Cauchy Sequences

A real sequence $\{a_n\}$ is convergent if and only if, for each $\epsilon >0$, there exists some integer $N$ such that

$$\underset{n\to \infty }{\lim }|a_n-a_m|=0\forall n,m\ge N$$

PROOF

TODO

NOTE

Sequences for which the above holds, i.e. convergent sequences, are also known as Cauchy sequences.

Properties

Theorem: Boundedness of Convergent Sequences

Every convergent real sequence is bounded.

PROOF

Suppose that $\{a_n\}$ converges to some $L\in \mathbb{R}$. Then, by definition, for each $\epsilon >0$, there exists some integer $N$ such that

$$|a_n-L|<\epsilon \forall n\ge N.$$

Choose $\epsilon =1$. The actual choice is irrelevant, it will just result in a different bound. Then,

$$|a_n-L|<1\forall n\ge N.$$

Let’s look at the absolute value of $a_n$:

$$|a_n|=|a_n-L+L|$$

Using the triangle inequality, we get

$$|a_n|=|a_n-L+L|\le |a_n-L|+|L|.$$

For all $n\ge N$,

$$|a_n-L|+|L|<1+|L|$$

and so $|a_n|<1+|L|$ for all $n\ge N$. This means that the modulus of all sequence terms from the $N$-th one onwards is less than $1+|L|$. Amongst the first $N-1$ terms of the sequence, choose the one whose modulus $M$ is greatest. The moduli of the first $N-1$ terms are thus all less than or equal to $M$. Essentially, we have

• $|a_n|\le M$ for every $n<N$;
• $|a_n|<1+|L|$ for every $n\ge N$.

Let $B=\max \{M,1+|L|\}$. Therefore, $|a_n|\le B$ for every integer $n$ and so $a_n$ is bounded.

Theorem: Convergence to Zero

A real sequence $\{a_n\}$ converges to zero if and only if $\{|a_n|\}$ converges to zero.

$$\underset{n\to \infty }{\lim }{a}_n=0⟺\underset{n\to \infty }{\lim }|a_n|=0$$

PROOF

TODO

The Squeeze Theorem for Sequences

Let $\{a_n\}$, $\{b_n\}$ and $\{c_n\}$ be Real Sequences such that both $\{a_n\}$ and $\{b_n\}$ converge to $L\in \mathbb{R}$.

If there exists an integer $N$ such that $a_n\le c_n\le b_n$ for all $n\ge N$, then $\{c_n\}$ also converges to $L$.

$$\underset{n\to \infty }{\lim }{c}_n=\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{b}_n=L$$

PROOF

Let $\epsilon >0$. Since $(a_n{)}_{n\in \mathbb{N}}$ and $(b_n{)}_{n\in \mathbb{N}}$ are convergent, there exist $N_a,N_b\in \mathbb{N}$ such that

$$|a_n-L|<\epsilon \forall n>N_a$$ $$|b_n-L|<\epsilon \forall n>N_b$$

We also assumed that there is an integer $N$ such that

$$a_n\le c_n\le b_n\forall n>N$$

It follows then

$$L-\epsilon <a_n\le c_n\le b_n<L+\epsilon \forall n\ge \max \{N,N_a,N_b\}$$

From this we see that

$$L-\epsilon \le c_n\le L+\epsilon \forall n\ge \max \{N,N_a,N_b\}$$

This is the same as

$$|c_n-L|<\epsilon n\ge \max \{N,N_a,N_b\}$$

Theorem: Limit Arithmetic

If $\{a_n\}$ and $\{b_n\}$ are both convergent Real Sequences, then

$$\begin{array}{rl}& \underset{n\to \infty }{\lim }(\alpha a_n+\beta b_n)=\alpha \underset{n\to \infty }{\lim }{a}_n+\beta \underset{n\to \infty }{\lim }{b}_n\forall \alpha ,\beta \in \mathbb{R}\\ & \\ & \underset{n\to \infty }{\lim }(a_n\cdot b_n)=\underset{n\to \infty }{\lim }{a}_n\cdot \underset{n\to \infty }{\lim }{b}_n\\ & \\ & \underset{n\to \infty }{\lim }\frac{a_n}{b_n}=\frac{\underset{n\to \infty }{\lim }{a}_n}{\underset{n\to \infty }{\lim }{b}_n},\underset{n\to \infty }{\lim }{b}_n\ne 0\\ & \\ & \underset{n\to \infty }{\lim }|a_n|=\left|\underset{n\to \infty }{\lim }{a}_n\right|\end{array}$$

PROOF

Let $A=\underset{n\to \infty }{\lim }{a}_n$ and $B=\underset{n\to \infty }{\lim }{b}_n$.

Proof of (1):

We have to prove that for each $\epsilon >0$, there exists some integer $N$ such that

$$|\alpha a_n+\beta b_n-(\alpha A+\beta B)|<\epsilon \forall n\ge N.$$

The case of $\alpha =\beta =0$ is trivial, since then $|\alpha a_n+\beta b_n-(\alpha A+\beta B)|$ is always zero and is thus smaller than all $\epsilon >0$. As for the existence of $N$ - not only does such an integer $N$ exist, but there are actually infinitely many such integers, since the inequality holds irrespective of $N$.

If $\alpha$ and $\beta$ are not zero, choose some arbitrary ${\epsilon }^{\prime }>0$. Since $a_n\to A$ and $b_n\to B$, there exist integers $N_A$ and $N_B$ such that

$$\begin{array}{r}|a_n-A|<{\epsilon }^{\prime }\forall n\ge N_A\\ |b_n-B|<{\epsilon }^{\prime }\forall n\ge N_B\end{array}$$

Let $N=\max \{N_A,N_B\}$. Then both inequalities hold for every $n\ge N$. Now, we look at

$$|\alpha a_n+\beta b_n-(\alpha A+\beta B)|=|\alpha (a_n-A)+\beta (b_n-B)|.$$

By the triangle inequality, we obtain

$$|\alpha (a_n-A)+\beta (b_n-B)|\le |\alpha (a_n-A)|+|\beta (b_n-B)|=|\alpha ||a_n-A|+|\beta ||b_n-B|.$$

For every $n\ge N$, we know that $|a_n-A|<{\epsilon }^{\prime }$ and $|b_n-B|<{\epsilon }^{\prime }$ and so we get

$$|\alpha ||a_n-A|+|\beta ||b_n-B|<|\alpha |{\epsilon }^{\prime }+|\beta |{\epsilon }^{\prime }.$$

Since $|\alpha (a_n-A)+\beta (b_n-B)|\le |\alpha ||a_n-A|+|\beta ||b_n-B|$ and $|\alpha a_n+\beta b_n-(\alpha A+\beta B)|=|\alpha (a_n-A)+\beta (b_n-B)|$, we get

$$|\alpha a_n+\beta b_n-(\alpha A+\beta B)|<|\alpha |{\epsilon }^{\prime }+|\beta |{\epsilon }^{\prime }\forall n\ge N$$

So far, the argument does not depend on the choice of ${\epsilon }^{\prime }$ which means that it must be true for all ${\epsilon }^{\prime }>0$. Thus, we have proven that for each $\epsilon =|\alpha |{\epsilon }^{\prime }+|\beta |{\epsilon }^{\prime }$, there exists some integer, namely $N$, such that

$$|\alpha a_n+\beta b_n-(\alpha A+\beta B)|<\epsilon \forall n\ge N,$$

which is what we set out to prove.

Proof of (2):

We have to prove that for each $\epsilon >0$, there exists some integer $N$ such that

$$|a_n\cdot b_n-A\cdot B|<\epsilon \forall n\ge N.$$

Choose some arbitrary ${\epsilon }^{\prime }>0$. Since $a_n\to A$ and $b_n\to B$, there exist integers $N_A$ and $N_B$ such that

$$\begin{array}{rl}& |a_n-A|<{\epsilon }^{\prime }\forall n\ge N_A\\ & |b_n-B|<{\epsilon }^{\prime }\forall n\ge N_B.\end{array}$$

Let $N=\max \{N_A,N_B\}$. Then both inequalities hold for all $n\ge N$. We transform the expression $|a_n\cdot b_n-A\cdot B|$ a bit:

$$|a_n\cdot b_n-A\cdot B|=|a_n\cdot b_n-A\cdot b_n+A\cdot b_n-A\cdot B|.$$

Using the triangle inequality, we obtain

$$|a_n\cdot b_n-A\cdot b_n+A\cdot b_n-A\cdot B|\le |a_n\cdot b_n-A\cdot b_n|+|A\cdot b_n-A\cdot B|,$$

which is equivalent to

$$|a_n\cdot b_n-A\cdot B|\le |b_n|\cdot |a_n-A|+|A|\cdot |b_n-B|.$$

For all $n\ge N$,

$$|b_n|\cdot |a_n-A|+|A|\cdot |b_n-B|<|b_n|\cdot {\epsilon }^{\prime }+|A|\cdot {\epsilon }^{\prime },$$

which in turn means that, for all $n\ge N$,

$$|a_n\cdot b_n-A\cdot B|<|b_n|\cdot {\epsilon }^{\prime }+|A|\cdot {\epsilon }^{\prime }$$

Recall that every convergent sequence is bounded. This means that there exists some $B>0$ and some integer $\mathcal{N}$ such that $|b_n|\le B$ for all $n\ge \mathcal{N}$. Therefore,

$$|a_n\cdot b_n-A\cdot B|<B\cdot {\epsilon }^{\prime }+|A|\cdot {\epsilon }^{\prime }\forall n\ge \mathcal{N}.$$

Set $\epsilon =B\cdot {\epsilon }^{\prime }+|A|\cdot {\epsilon }^{\prime }$. Therefore,

$$|a_n\cdot b_n-A\cdot B|<\epsilon \forall n\ge \mathcal{N}.$$

It is obvious that $\epsilon$ can be any positive real number. Moreover, since the argument so far does not depend on any particular choice of $\epsilon$, we know the argument holds for all $\epsilon >0$. We have thus proven that for each $\epsilon >0$, there exists some integer, namely $\mathcal{N}$, such that

$$|a_n\cdot b_n-A\cdot B|<\epsilon \forall n\ge \mathcal{N},$$

which is what we set out to show.

Proof of (3):

TODO

Proof of (4):

TODO

Divergence and Infinite Limits

Definition: Divergence of Real Sequences

A real sequence is divergent iff it does not converge to any $\mathbb{R}$.

There are two special types of divergence in which we are often interested.

Definition: Divergence towards Positive Infinity

A real sequence $\{a_n\}$ diverges towards positive infinity iff for each $A\in \mathbb{R}$ there is some integer $N$ such that

$$a_n>A\forall n\ge N$$

NOTATION

$$\underset{n\to \infty }{\lim }{a}_n=\infty$$

Definition: Divergence towards Negative Infinity

A real sequence $\{a_n\}$ diverges towards negative infinity iff for each $A\in \mathbb{R}$ there is some integer $N$ such that

$$a_n<A\forall n\ge N$$

NOTATION

$$\underset{n\to \infty }{\lim }{a}_n=-\infty$$

Even though we use limit notation for sequences that diverge towards positive or negative infinity, these sequences are not convergent and their limits do not exist. However, we often talk of “infinite limits” because of the notation we have chosen. Just remember that, strictly speaking, the “limits” of divergent Real Sequences never exist.

Properties

Theorem: The Limit of $q^n$

The real sequence $q^n$:

• converges to $0$ if and only if $|q|<1$;

• converges to $1$ if and only if $|q|=1$;

• diverges if and only if $|q|>1$;

• diverges towards $+\infty$ if and only if $q>1$.

PROOF

TODO

Theorem: The Limit of $\frac{1}{n^k}$

For every natural number $k>0$, the real sequence $\{\frac{1}{n^k}\}$ converges to $0$.

$$\underset{n\to \infty }{\lim }\frac{1}{n^k}=0$$

PROOF

The following proof was generated by AI and may contain mistakes. TODO: Review

We want to prove that the sequence $a_n=\frac{1}{n^k}$ converges to $0$ for any fixed natural number $k>0$. By the definition of convergence, we need to show that for every $ϵ>0$, there exists a natural number $N$ such that for all $n>N$, we have $|a_n-0|<ϵ$.

Let $ϵ>0$ be an arbitrary positive real number. We want to find an $N\in \mathbb{N}$ such that for all $n>N$,

$|\frac{1}{n^k}-0|<ϵ$

Since $n$ is a natural number, $n\ge 1$. Since $k$ is a natural number $k>0$, we have $n^k\ge 1^k=1$. Thus, $n^k$ is always positive. Therefore, the absolute value simplifies to:

$|\frac{1}{n^k}-0|=|\frac{1}{n^k}|=\frac{1}{n^k}$

So, we need to find $N\in \mathbb{N}$ such that for all $n>N$,

$\frac{1}{n^k}<ϵ$

Since $ϵ>0$ and $n^k>0$, we can rearrange the inequality: Take the reciprocal of both sides (this reverses the inequality sign):

$n^k>\frac{1}{ϵ}$

Since $k>0$, we can take the $k$-th root of both sides. The function $x\mapsto x^{1/k}$ is strictly increasing for positive $x$.

$n>{\left(\frac{1}{ϵ}\right)}^{1/k}$

We need to find a natural number $N$ such that for all $n>N$, the condition $n>(\frac{1}{ϵ}{)}^{1/k}$ holds. By the Archimedean Property of the real numbers, for any positive real number, such as $(\frac{1}{ϵ}{)}^{1/k}$, there exists a natural number $N$ greater than it. So, we can choose $N$ to be any natural number such that

$N>{\left(\frac{1}{ϵ}\right)}^{1/k}$

For example, we can choose $N=\lfloor (\frac{1}{ϵ}{)}^{1/k}\rfloor +1$.

Now, let’s verify this choice of $N$. Suppose $n$ is any natural number such that $n>N$. Since $N>(\frac{1}{ϵ}{)}^{1/k}$, we have $n>N>(\frac{1}{ϵ}{)}^{1/k}$. So, $n>(\frac{1}{ϵ}{)}^{1/k}$. Since $k>0$, raising both sides to the power of $k$ (which is an increasing function for positive values) preserves the inequality:

$n^k>{\left({\left(\frac{1}{ϵ}\right)}^{1/k}\right)}^k$

$n^k>\frac{1}{ϵ}$

Since $n^k>0$ and $ϵ>0$, we can take the reciprocal of both sides, which reverses the inequality:

$\frac{1}{n^k}<ϵ$

Since we already established that $|\frac{1}{n^k}-0|=\frac{1}{n^k}$, we have shown that for any $n>N$,

$|\frac{1}{n^k}-0|<ϵ$

Since we found such an $N$ for an arbitrary $ϵ>0$, by the definition of convergence, the sequence $\frac{1}{n^k}$ converges to $0$.

This completes the proof.

Theorem: Reciprocal Limits

Let $\{a_n\}$ be a real sequence.

If $\{a_n\}$ converges to $0$ and there exists some integer $N$ such that $a_n>0$ for all $n\ge N$, then $\{\frac{1}{a_n}\}$ diverges towards $+\infty$.

$$a_n>0\text{ and }\underset{n\to \infty }{\lim }{a}_n=0⟹\underset{n\to \infty }{\lim }\frac{1}{a_n}=+\infty$$

If $\{a_n\}$ converges to $0$ and there exists some integer $N$ such that $a_n<0$ for all $n\ge N$, then $\{\frac{1}{a_n}\}$ diverges towards $-\infty$.

$$a_n<0\text{ and }\underset{n\to \infty }{\lim }{a}_n=0⟹\underset{n\to \infty }{\lim }\frac{1}{a_n}=-\infty$$

If $\{a_n\}$ diverges towards either $+\infty$ or $-\infty$, then $\{\frac{1}{a_n}\}$ converges to $0$.

$$\underset{n\to \infty }{\lim }{a}_n=\pm \infty ⟹\underset{n\to \infty }{\lim }\frac{1}{a_n}=0$$

PROOF

TODO

Theorem: Arithmetic with Infinite Limits

Let $\{a_n\}$ and $\{b_n\}$ be Real Sequences.

If $\{a_n\}$ converges but $\{b_n\}$ diverges towards $\pm \infty$, then $\{a_n+b_n\}$ also diverges towards $\pm \infty$.

$$\underset{n\to \infty }{\lim }{a}_n\in \mathbb{R}\text{ and }\underset{n\to \infty }{\lim }{b}_n=\pm \infty ⟹\underset{n\to \infty }{\lim }(a_n+b_n)=\pm \infty$$

If $\{a_n\}$ converges towards $L\in \mathbb{R}$ but $\{b_n\}$ diverges towards $\pm \infty$, then $\{a_n\cdot b_n\}$ also diverges towards $\pm \infty$ when $L>0$ and diverges towards $\mp \infty$ when $L<0$.

$$\underset{n\to \infty }{\lim }{a}_n=L\in \mathbb{R}\text{ and }\underset{n\to \infty }{\lim }{b}_n=\pm \infty ⟹\underset{n\to \infty }{\lim }(a_n\cdot b_n)=\{\begin{array}{l}\pm \infty \text{ if }L>0\\ \mp \infty \text{ if }L<0\end{array}$$

If $\{a_n\}$ and $\{b_n\}$ both diverge towards $\pm \infty$, then $\{a_n+b_n\}$ also diverges towards $\pm \infty$ and $\{a_n\cdot b_n\}$ diverges towards $+\infty$.

$$\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }{b}_n=\pm \infty ⟹\underset{n\to \infty }{\lim }(a_n+b_n)=\pm \infty \text{ and }\underset{n\to \infty }{\lim }(a_n\cdot b_n)=+\infty$$

If $\{a_n\}$ diverges towards $\pm \infty$ but $\{b_n\}$ diverges towards $\mp \infty$, then $\{a_n\cdot b_n\}$ diverges towards $-\infty$. However, this information is insufficient to determine ${\lim }_{n\to \infty }(a_n+b_n)$.

$$\underset{n\to \infty }{\lim }{a}_n=\pm \infty \text{ and }\underset{n\to \infty }{\lim }{b}_n=\mp \infty ⟹\underset{n\to \infty }{\lim }(a_n\cdot b_n)=-\infty$$

PROOF

TODO

Theorem: The Limit of ${\left(1+\frac{1}{n}\right)}^n$ and its Variations

The real sequence $a_n={\left(1+\frac{1}{n}\right)}^n$ converges towards Euler’s number $\mathrm{e}$.

$$\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n=\mathrm{e}$$

Similarly, the limits of the following Real Sequences can also be expressed using the real exponential function:

$$\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^{n+r}=\mathrm{e}\forall r\in \mathbb{R}$$ $$\underset{n\to \infty }{\lim }{\left(1+\frac{r}{n}\right)}^n={\mathrm{e}}^r\forall r\in \mathbb{R}$$

PROOF

TODO