Home Calculus Convergence of sequences

# Convergence of sequences

We discuss the convergence of sequences and how to calculate the limit of a sequence. This subject is fundamental in real analysis because many proofs of theorems rely on the convergence of an appropriate sequence.

We also offer several exercises with detailed solutions to make good use of the material in this course. This course is for first-year students and part of the article may also relate to the final year of the high school math curriculum.

## Convergence of sequences: definition and properties

We assume that the reader knows the real numbers and the distance between the numbers.

A sequence of real numbers is an application $u:\mathbb{N}\to \mathbb{R}$, such that for any $n\in\mathbb{R}$ we associate a real number $u(n)\in\mathbb{R}$. As usual, we use the notation $u(n):=u_n$, and the sequence will be denoted by $(u_n)_n$. For example, we select\begin{align*}u_n=\frac{1}{n},\quad n\ge 1.\end{align*} Note that we can also speak of sequences of complex numbers if we replace in the above definition the set of real numbers with the set of complex numbers $\mathbb{C}$.

Convergence to a real number: As a matter of fact, it is not easy to give a rigorous definition of the convergence of sequences for beginners. Here we will give a simple definition.

We say that the sequence $(u_n)_n$ converges to a real number $\ell$ if we can find a range $N\in\mathbb{N}$ such that all terms $u_n$ for $n\ge N$ are closed to $\ell$. This means that the distance between $u_n$ and $\ell$ is small enough whenever $n\ge N$. Mathematically, this distance is exactly $|u_n-\ell|$. More precisely this means that or any small real number $\varepsilon$ we have $|u_n-\ell|\le \varepsilon$ whenever the positive integer $n\ge N$.

Summary: the sequnce $(u_n)_n$ converges to $\ell$ if and only if for any small $\varepsilon>0,$ there existe a suffiently large $N\in\mathbb{N}$ such that for any $n\in \mathbb{N}$ with $n\ge N,$ we have $|u_n-\ell|\le \varepsilon$.

When the sequence $(u_n)_n$ converges to $\ell$, we say that $\ell$ is the limit of the sequence, and we write\begin{align*}\ell=\lim_{n\to+\infty}u_n.\end{align*}

The question is how to determine $N$. Let us show that the sequence $\frac{1}{n}$ converges to $0$. According to the discussion above, the distance $|\frac{1}{n}-0|= \frac{1}{n}$ should be small enough. This means that if we take a very small number $\varepsilon>0$ we will have $\frac{1}{n}< \varepsilon$. So $n>\frac{1}{\varepsilon}$. We then consider that $N$ is the smallest natural number greater than $\frac{1}{\varepsilon}$.

Remarque: if the limit of a sequence exists then it is unique.

A geometric sequence: Let $u_n=a^n$ for any $n\in\mathbb{N}$, where $a\in\mathbb{R}$. Let us prove that when $a\in (-1,1)$, the geometric sequence converges to $0$. In fact, by the same arguments as above, we take an arbitrary very small real $\varepsilon\in (0,1)$ such that $|a|^n<\varepsilon$. An the function $x\mapsto \ln(x)$ is increasing, we have $n\ln(|a|)<\ln(\varepsilon)$. On the other hand, as $|a|$ and $\varepsilon$ are in $(0,1)$, then their logarithm is negative. Thus \begin{align*} n>\frac{ \ln(\varepsilon) }{ \ln(|a|) }.\end{align*} We choose $N$ the smallest natural number such that $N> \ln(\varepsilon)/ \ln(|a|)$. Hence \begin{align*}\lim_{n\to+\infty} a^n=0\end{align*}for any $a\in (-1,1)$.

Convergence to $+\infty$: A sequence converges to $+\infty$ if: for any sufficiently large real number $A>0$, there exists an integer sufficiently large number $N\in\mathbb{N}$ such for any $n\in\mathbb{N},$ $n\ge N$ implies that $u_n>A$.

Convergence to $-\infty$: A sequence converges to $-\infty$ if: for any sufficiently large real number $A>0$, there exists an integer sufficiently large number $N\in\mathbb{N}$ such for any $n\in\mathbb{N},$ $n\ge N$ implies that $u_n<-A$.

Divergence sequences: We say that a sequence is divergent if its limit is equal to $\pm \infty,$ or the limit does not exist at all.

Proposition: Every convegrente sequence $(u_n)_n$ is bounded, i.e. there exists a real number $M>0$ such that $|u_n|\le M$ for any $n\in\mathbb{N}$.

Proof: Assume that $(u_n)_n$ converges to $\ell\in\mathbb{R}$. Thus for $\varepsilon=1,$ there exists $N\in \mathbb{N}$ such that $|u_n-\ell|\le 1$ for any $n\ge N$, so that $$|u_n|=|(u_n-\ell)+\ell|\le |u_n-\ell|+|\ell|\le 1+|\ell|$$ for any $n\ge N$. We set $\delta:=\max\{|u_0|,\cdots,|u_{N-1}|\}$. Now if we select $M:=\max\{1+|\ell|,\delta\}$, we obtain $|u_n|\le M$ for any $n\in\mathbb{N}$.

## How to calculate the limit of a sequence?

In this paragraph, we provide techniques to prove the convergence of sequences and in some cases even calculate the exact limit of the sequence. We first start with some definitions:

Increasing and decreasing: A sequence $(u_n)_n$ is increasing if $u_{n+1}\ge u_n$ for all $n\in\mathbb{N}$. It is decreasing if $u_{n+1}\le u_n$ for all $n$. A sequence is said to be monotone if it is increasing or decreasing. We can also speak of a strictly monotone sequence if we replace respectively “$\le$” and “$\ge$” by “$<$” and “$>$”.

Proposition: Let $(u_n)$ be a real sequence. The following assertions hold:

• If $(u_n)$ is increasing and there exists a real number $M>0$ such that $u_n\le M$ for any $n,$ the the sequence $(u_n)_m$ is convergente.
• The sequence $(u_n)_n$ is convergent if it is decreasing and there exists a real number $m>0$ such that $u_n\ge m$ for any $n$.

This proposition provides us only information about the convergence of the sequences but not about the exact limit. The following result gives at the same time the convergence and the value of the limit.

The Squeeze theorem: Let three sequences $(u_n)_n,(v_n)_n$ and $(w_n)_n$ such that\begin{align*}&v_n\le u_n\le w_n,\quad\text{for all}\;n\in\mathbb{N},\quad\text{and}\cr & \lim_{n\to+\infty}w_n=\lim_{n\to+\infty}v_n=\ell,\end{align*}then the sequence $(u_n)_n$ also converges to the same limit $\ell$.

A practical consequence of this theorem is: if there exists a positive sequence $(\alpha_n)_n$ converging to $0$ and if a sequence $(u_n)_n$ satisfies $|u_n|\le\alpha_n$, then $(u_n)_n$ converges also to $0$. This is because $-\alpha_n\le u_n\le \alpha_n$ for all $n$.

The use of continuous functions: Many sequences are of the form $u_n=f(v_n),$ where $f$ is a continuous function and $(v_n)_n$ is a convergence sequence that we know its limit $\ell$. Thus $(u_n)_n$ is also a convergent sequence and its limit is exactly $f(\ell)$. For example $u_n=\sin(\frac{1}{n})$. We know that $v_n=\frac{1}{n}\to 0$ and the function $f(x)=\sin(x)$ is continuous on $\mathbb{R}$. Then $(u_n)_n$ converges to $f(0)=\sin(0)=0$.

## Exercises on sequences

In the following examples, we show that the computation of the limit of some sequences strongly depends on the squeeze theorem, on the geometric sequences, and on the fact that the sequence $(\frac{1}{n})_n$ converges to $0$.

Exercise: Determine the limit of the sequence $u_n=\frac{\sin(n^2)}{n}$ for natural numbers $n\ge 1$.

Proof: We know that the sinus of any real number belongs to $[-1,1]$. Using this fact, we have $|\sin(n^2)|\le 1$. Then \begin{align*}|u_n|\le \frac{1}{n}.\end{align*} Using the background of the previous paragraph, we deduce that the sequence $(u_n)_n$ converges to $0$.

Exercise: Compute the limit of the sequence $v_n=2^n\sin(\frac{1}{3^n})$ for $n\in\mathbb{N}$.

Proof: We recall that $|\sin(x)|\le |x|$ for any real number $x,$. Applying this inequality, we obtain \begin{align*}|v_n|\le \left(\frac{2}{3}\right)^n.\end{align*} On the other hand, as $\frac{2}{3}\in (-1,1)$, then the geometric sequence $((\frac{2}{3})^n)_n$ converges to $0$. Thus the sequence $(v_n)_n$ converges to $0$.

Exercise: Calculate the limit of the sequence $u_n=\arctan(\frac{2^n-3^n}{2^n+3^n})$.

Proof: We select $f(x)=\arctan(x)$ for $x\in\mathbb{R}$ and $v_n=\frac{2^n-3^n}{2^n+3^n}$. Then $u_n=f(v_n)$. As $f$ is continuous on $\mathbb{R},$ to compute the limit of the sequence $(u_n)$, it suffices to determine that of the sequence $(v_n)_n$. Observe that \begin{align*} v_n&= \frac{2^n \left(1- \left( \frac{2}{3}\right)^n\right)}{2^n \left(1+ \left( \frac{2}{3}\right)^n\right)}\cr &=\frac{1- \left( \frac{2}{3}\right)^n}{1+ \left( \frac{2}{3}\right)^n}.\end{align*} As the geometric sequence $((\frac{2}{3})^n)_n$ converges to $0$. then $(v_n)_n$ converges to $1$. Hence $(u_n)_n$ converges to $f(1)=\arctan(1)=\frac{\pi}{4}$.

Exercise: Prove that any real number is a limit of a sequence in $\mathbb{Q},$ the rational numbers set.

Proof: Here we will use the squeeze theorem. In fact, let $x\in\mathbb{R}$. We construct a sequence, of course depending on $x$, of elements in $\mathbb{Q}$ that converges to $x$. We select \begin{align*} u_n=\frac{[nx]}{n},\quad n\in \mathbb{N}^\ast,\end{align*} where $[nx]$ is the integer part of the real number $nx$. It satisfies $nx-1<[nx]\le nx$. Thus, by dividing by $n$ for any $n\in \mathbb{N}^\ast,$  we obtain $$x-\frac{1}{n}<u_n\le x.$$ Now by the squeeze theorem, the sequence $(u_n)_n$ converges to $x$. The result follows now because $u_n\in\mathbb{Q}$ for any $n$.

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