How to prove convergence for recursive sequences – "Math for Non-Geeks"

This article shows you how to prove convergence for recursively defined sequences, i.e. if only a recurrence $a_{n+1}=f(a_n)$ and an initial value are given, and it is not known what $a_n$ explicitly looks like. The monotonicity criterion will turn out very useful for this.

Problem setting

We want to solve problems of the following kind:

Let $(a_n{)}_{n\in {\mathbb{N}}_0}$ be a sequence defined by $a_0=0$ and $a_{n+1}=\frac{1}{2}{a}_n+\frac{1}{2}$ for $n\in {\mathbb{N}}_0$ . Does it converge? If yes, what is its limit value?

Applying the epsilon-criterion is complicated, here: it is not a priori known what elements with large indices look like (e.g. $a_{1000}$), which makes it hard to guess a limit $a$. It is even harder to give a bound for $|a-a_n|$, as the explicit form of $a_n$ is often not known.

Problem-solving strategies

The article will cover two strategies for proving convergence of a recursively defined sequence:

1. Finding the explicit form: You may try to find the explicit form of the sequence elements $a_n$. Then it can be easier to determine a limit $a$ and prove convergence.
2. Using the monotonicity criterion: If you can not find an explicit form, but you are convinced that the sequence should converge, you may try to use this criterion. It works without knowing an explicit form.

Finding the explicit form

One way to prove convergence is to try to find an explicit form for the sequence elements $a_n$ . It is useful to compute the first sequence elements in order to get a feeling of how the sequence behaves. In the above example:

$\begin{array}{cc}{a}_0& =0\\ a_1& =\frac{1}{2}{a}_0+\frac{1}{2}=\frac{1}{2}\cdot 0+\frac{1}{2}=\frac{1}{2}\\ a_2& =\frac{1}{2}{a}_1+\frac{1}{2}=\frac{1}{2}\cdot \frac{1}{2}+\frac{1}{2}=\frac{3}{4}\\ a_3& =\frac{1}{2}{a}_2+\frac{1}{2}=\frac{1}{2}\cdot \frac{3}{4}+\frac{1}{2}=\frac{7}{8}\\ a_4& =\frac{1}{2}{a}_3+\frac{1}{2}=\frac{1}{2}\cdot \frac{7}{8}+\frac{1}{2}=\frac{15}{16}\\ & ⋮\end{array}$

These elements are all smaller than 1, while slowly approaching it. We therefore conjecture that $a_n$ converges to 1. Let us try to find an explicit form, i.e. a map $n\mapsto a_n$. That means, we are looking for a term $f(x)$ with:

$\begin{array}{cc}f(0)& =0\\ f(1)& =\frac{1}{2}\\ f(2)& =\frac{3}{4}\\ f(3)& =\frac{7}{8}\\ & ⋮\end{array}$

The denominator suspiciously looks like a power of $2$ . The numerator seems to be always one less than the denominator: $\begin{array}{c}\begin{array}{ccc}f(0)& =0& =\frac{2^0-1}{2^0}\\ f(1)& =\frac{1}{2}& =\frac{2^1-1}{2^1}\\ f(2)& =\frac{3}{4}& =\frac{2^2-1}{2^2}\\ f(3)& =\frac{7}{8}& =\frac{2^3-1}{2^3}\\ & ⋮\end{array}\end{array}$

That would mean, $f(n)=\frac{2^n-1}{2^n}$ . So we assert that

$a_n=f(n)=\frac{2^n-1}{2^n}=1-\frac{1}{2^n}=1-{\left(\frac{1}{2}\right)}^n$

Now, we should try to prove or disprove this assertion. Induction is a suitable way to do this, as the recursion relation $a_{n+1}=\frac{1}{2}{a}_n+\frac{1}{2}$ may perfectly serve as an induction step. The induction starts with $a_0=0=1-1=1-{\left(\frac{1}{2}\right)}^0$

The induction step from $n$ to $n+1$ is done using the recursion relation:

$\begin{array}{cc}{a}_{n+1}& =\frac{1}{2}{a}_n+\frac{1}{2}\\ & \downarrow \text{induction assumption}\\ & =\frac{1}{2}\cdot (1-{\left(\frac{1}{2}\right)}^n)+\frac{1}{2}\\ & =\frac{1}{2}-{\left(\frac{1}{2}\right)}^{n+1}+\frac{1}{2}\\ & =1-{\left(\frac{1}{2}\right)}^{n+1}\end{array}$

So we indeed proved $a_n=1-{\left(\frac{1}{2}\right)}^n$ . This explicit form allows us to use the usual tools (from the previous articles) for proving convergence. For instance, we know $\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{1}{2}\right)}^n=0$ and we can use the limit theorems:

$\begin{array}{cc}\underset{n\to \mathrm{\infty }}{lim}{a}_n& =\underset{n\to \mathrm{\infty }}{lim}1-{\left(\frac{1}{2}\right)}^n\\ & \downarrow \text{Limit theorems: }\underset{n\to \mathrm{\infty }}{lim}(a_n+b_n)=\underset{n\to \mathrm{\infty }}{lim}{a}_n+\underset{n\to \mathrm{\infty }}{lim}{b}_n\\ & =\underset{n\to \mathrm{\infty }}{lim}1-\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{1}{2}\right)}^n\\ & =1-0=1\end{array}$

Alternative way: using a telescoping sum

Sometimes, finding a simple explicit form may be enormously hard or even impossible. In any case, one can write the elements $a_n$ using telescoping sums. To illustrate this, we consider:

$a_0=0,a_1=1,a_n=\frac{1}{2}{a}_{n-1}+\frac{1}{2}{a}_{n-2}(n\ge 2)$

The difference between two sequence elements $a_n$ and $a_{n-1}$ is

$\begin{array}{cc}{a}_n-a_{n-1}& =\frac{1}{2}{a}_{n-1}+\frac{1}{2}{a}_{n-2}-a_{n-1}\\ & =\frac{1}{2}{a}_{n-2}-\frac{1}{2}{a}_{n-1}\\ & =\frac{1}{2}(a_{n-2}-a_{n-1})\\ & =-\frac{1}{2}(a_{n-1}-a_{n-2})\end{array}$

For $n\ge 3$, applying this step again yields

$\begin{array}{cc}{a}_n-a_{n-1}& =-\frac{1}{2}(a_{n-1}-a_{n-2})\\ & =-\frac{1}{2}(\frac{1}{2}{a}_{n-2}+\frac{1}{2}{a}_{n-3}-a_{n-2})\\ & =-\frac{1}{2}(\frac{1}{2}{a}_{n-3}-\frac{1}{2}{a}_{n-2})\\ & ={(-\frac{1}{2})}^2(a_{n-2}-a_{n-3})\end{array}$

So each time, we get an additional factor of $-\frac{1}{2}$ . After $n-1$ steps, there will be

$a_n-a_{n-1}={(-\frac{1}{2})}^{n-1}\underset{=1}{\underbrace{(a_1-a_0)}}={(-\frac{1}{2})}^{n-1}$

This is not yet an explicit form for $a_n$, but for the differences $a_n-a_{n-1}$. However, we can express $a_n$ using telescoping sums:

$\sum _{k=1}^n(a_k-a_{k-1})=a_n-a_0=a_n$

Substituting the formula for the differences gives a geometric sum

$\sum _{k=1}^n{(-\frac{1}{2})}^{k-1}=\sum _{k=0}^{n-1}{(-\frac{1}{2})}^k=\frac{1-(-\frac{1}{2}{)}^n}{1-(-\frac{1}{2})}=\frac{1-(-\frac{1}{2}{)}^n}{\frac{3}{2}}=\frac{2}{3}(1-(-\frac{1}{2}{)}^n)$

This allows for an explicit expression

$a_n=\frac{2}{3}(1-(-\frac{1}{2}{)}^n)$

Using $\underset{n\to \mathrm{\infty }}{lim}(-\frac{1}{2}{)}^n=0$ and the limit theorems, we get

$\underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{2}{3}(1-(-\frac{1}{2}{)}^n)=\frac{2}{3}(1-0)=\frac{2}{3}$

Using the monotonicity criterion

Step 1: Proving convergence

What if we cannot find an explicit form of the sequence elements? If the sequence is monotonically increasing or decreasing, the monotonicity criterion can be useful. We recall this criterion:

All bounded and monotone real sequences converge.

Hence, it suffices to show boundedness and monotonicity of the sequence. For instance, returning to the initial example $a_0=0$ and $a_{n+1}=\frac{1}{2}{a}_n+\frac{1}{2}$:

$\begin{array}{cc}{a}_0& =0\\ a_1& =\frac{1}{2}{a}_0+\frac{1}{2}=\frac{1}{2}\cdot 0+\frac{1}{2}=\frac{1}{2}\\ a_2& =\frac{1}{2}{a}_1+\frac{1}{2}=\frac{1}{2}\cdot \frac{1}{2}+\frac{1}{2}=\frac{3}{4}\\ a_3& =\frac{1}{2}{a}_2+\frac{1}{2}=\frac{1}{2}\cdot \frac{3}{4}+\frac{1}{2}=\frac{7}{8}\\ a_4& =\frac{1}{2}{a}_3+\frac{1}{2}=\frac{1}{2}\cdot \frac{7}{8}+\frac{1}{2}=\frac{15}{16}\\ & ⋮\end{array}$

This seems to be monotonically increasing and bounded by $1$.

Monotonicity is proven inductively. Clearly $a_1\ge a_0$, as

$a_1=\frac{1}{2}{a}_0+\frac{1}{2}=\frac{1}{2}\cdot 0+\frac{1}{2}=\frac{1}{2}\ge 0=a_0$

The induction step from $n$ to $n+1$ consists of showing that the sequence element gets bigger within that step:

$\begin{array}{cc}{a}_{n+1}& =\frac{1}{2}{a}_n+\frac{1}{2}\\ & \downarrow \text{induction assumption: }{a}_n\ge a_{n-1}\\ & \ge \frac{1}{2}{a}_{n-1}+\frac{1}{2}=a_n\end{array}$

This establishes monotonicity of the sequence. Boundedness by $1$ can also be shown inductively. Of course, $a_0=0$ which is smaller than 1. In the induction step, we need to show that if $a_n\le 1$ , then also $a_{n+1}\le 1$:

$\begin{array}{cc}{a}_{n+1}& =\frac{1}{2}{a}_n+\frac{1}{2}\\ & \downarrow \text{induction assumption: }{a}_n\le 1\\ & \le \frac{1}{2}1+\frac{1}{2}=1\end{array}$

Hence, $(a_n{)}_{n\in {\mathbb{N}}_0}$ is bounded from above by $1$ . Now, the monotonicity criterion can be applied: The sequence is monotonically increasing and bounded from above, so it must converge.

Step 2: finding the limit

The convergence above can be used for finding the limit. By step 1, we already know that there must be a limit $a\in ℝ$ with $\underset{n\to \mathrm{\infty }}{lim}{a}_n=a$ and it can be at most the upper bound, namely 1. As the sequence elements approach 1 closer and closer, we assert that it is exactly 1.

To verify this, we take a look at $a=\underset{n\to \mathrm{\infty }}{lim}{a}_{n+1}$:

$\begin{array}{cc}a& =\underset{n\to \mathrm{\infty }}{lim}{a}_{n+1}\\ & \downarrow \text{recursion relation: }{a}_{n+1}=\frac{1}{2}{a}_n+\frac{1}{2}\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2}{a}_n+\frac{1}{2}\\ & \downarrow \text{limit theorems}\\ & =\left(\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2}\right)\cdot \left(\underset{n\to \mathrm{\infty }}{lim}{a}_n\right)+\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2}\\ & =\frac{1}{2}a+\frac{1}{2}\end{array}$

So if there is any $a$ , it must fulfil $a=\frac{1}{2}a+\frac{1}{2}$ . We resolve the equation for $a$ and get

$a=\frac{1}{2}a+\frac{1}{2}⟺\frac{1}{2}a=\frac{1}{2}⟺a=1$

So $a=1$ is indeed the limit of the sequence $(a_n{)}_{n\in {\mathbb{N}}_0}$.

Question: Above, we used that if $\underset{n\to \mathrm{\infty }}{lim}{a}_n=a$ is known, then there is also $\underset{n\to \mathrm{\infty }}{lim}{a}_{n+1}=a$ . Why is this the case?

$\underset{n\to \mathrm{\infty }}{lim}{a}_{n+1}$ is the limit of the sequence $(a_{n+1}{)}_{n\in {\mathbb{N}}_0}$. This is just a subsequence of $(a_n{)}_{n\in {\mathbb{N}}_0}$, which contains all but the first element. Since each subsequence converges to the same limit as the original one, we have $\underset{n\to \mathrm{\infty }}{lim}{a}_{n+1}=\underset{n\to \mathrm{\infty }}{lim}{a}_n=a$.

Exercises

Exercise 1

Exercise (Finding the limit of a recursively defined sequence)

Let $a_n$ be recursively defined by:

$a_0=2,a_{n+1}=2-\frac{1}{a_n}$

Prove that this sequence converges and compute the limit by

1. finding an explicit form of the sequence
2. using the monotonicity criterion

Solution (Finding the limit of a recursively defined sequence)

Solution sub-exercise 1:

There is

$\begin{array}{cc}{a}_0& =2\\ a_1& =2-\frac{1}{2}=\frac{3}{2}\\ a_2& =2-\frac{1}{\frac{3}{2}}=2-\frac{2}{3}=\frac{4}{3}\\ a_3& =2-\frac{1}{\frac{4}{3}}=2-\frac{3}{4}=\frac{5}{4}\\ & ⋮\end{array}$

So we assert that $a_n=\frac{n+2}{n+1}$ for all $n\in {\mathbb{N}}_0$. The proof goes by induction on $n$:

Theorem whose validity shall be proven for $n\in {\mathbb{N}}_0$ bewiesen werden soll:

$a_n=\frac{n+2}{n+1}$

1. Base case:

For $n=0$ we have $a_0=\frac{0+2}{0+1}=2$.

2. Inductive step:

2a. Inductive hypothesis:

$a_n=\frac{n+2}{n+1}$

2b. Induction theorem:

$a_{n+1}=\frac{n+3}{n+2}$

2c. Proof of induction step:

$\begin{array}{cc}{a}_{n+1}& =2-\frac{1}{a_n}\stackrel{\text{IH}}{=}2-\frac{1}{\frac{n+2}{n+1}}\\ & =2-\frac{n+1}{n+2}=\frac{2n+4}{n+2}-\frac{n+1}{n+2}=\frac{n+3}{n+2}\end{array}$

So we have an explicit form. The limit theorems now directly yield the solution

$\underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{n+2}{n+1}=\underset{n\to \mathrm{\infty }}{lim}\frac{1+\frac{2}{n}}{1+\frac{1}{n}}=\frac{1}{1}=1$

Solution sub-exercise 2:

The lower-bound induction below proves $a_n\ge 1$ for all $n\in {\mathbb{N}}_0$, using only the recurrence and the initial value. This also shows that the recurrence is well-defined and all its denominators are positive. Using this fact, we show that $(a_n{)}_{n\in {\mathbb{N}}_0}$ is monotonically decreasing, i.e. $a_{n+1}\le a_n$ for all $n\in {\mathbb{N}}_0$. This is done by induction over $n$:

Theorem whose validity shall be proven for $n\in {\mathbb{N}}_0$ bewiesen werden soll:

$a_{n+1}\le a_n$

1. Base case:

$a_1=2-\frac{1}{a_0}=2-\frac{1}{2}=\frac{3}{2}\le 2=a_0$

2. Inductive step:

2a. Inductive hypothesis:

$a_{n+1}\le a_n$

2b. Induction theorem:

$a_{n+2}\le a_{n+1}$

2c. Proof of induction step:

$a_{n+2}=2-\frac{1}{a_{n+1}}\stackrel{\text{IH}}{\le }2-\frac{1}{a_n}=a_{n+1}$

In addition, $(a_n{)}_{n\in {\mathbb{N}}_0}$ is bounded from below by $1$ , i.e. $a_n\ge 1$ for all $n\in {\mathbb{N}}_0$. This is also shown using induction:

Theorem whose validity shall be proven for $n\in {\mathbb{N}}_0$ bewiesen werden soll:

$a_n\ge 1$

1. Base case:

$a_0=2\ge 1$

2. Inductive step:

2a. Inductive hypothesis:

$a_n\ge 1$

2b. Induction theorem:

$a_{n+1}\ge 1$

2c. Proof of induction step:

$a_{n+1}=2-\frac{1}{a_n}\stackrel{\text{IH}}{\ge }2-\frac{1}{1}=2-1=1$

The monotonicity criterion then yields convergence of $(a_n{)}_{n\in {\mathbb{N}}_0}$ and there is $\underset{n\to \mathrm{\infty }}{lim}{a}_n=a=\underset{n\to \mathrm{\infty }}{lim}{a}_{n+1}$. The lower bound implies $a\ge 1$, so the quotient rule applies. We can get the limit by taking the recursion relation and resolving it for $a$:

$\begin{array}{cc}& \underset{=a}{\underbrace{\underset{n\to \mathrm{\infty }}{lim}{a}_n}}=\underset{n\to \mathrm{\infty }}{lim}2-\frac{1}{a_n}=2-\frac{1}{\underset{n\to \mathrm{\infty }}{lim}{a}_n}=2-\frac{1}{a}\\ ⟺& a^2=2a-1\\ ⟺& \underset{=(a-1{)}^2}{\underbrace{a^2-2a+1}}=0\\ ⟺& a=1\end{array}$

Hence, $(a_n{)}_{n\in {\mathbb{N}}_0}$ converges to $a=1$.

Exercise 2

Exercise (Proving convergence of a recursively defined sequence)

We consider the following recursive sequence

$\begin{array}{cc}{a}_0& =0\\ a_1& =2\\ a_n& =\frac{1}{3}{a}_{n-1}+\frac{2}{3}{a}_{n-2}(n\ge 2)\end{array}$

1. Prove that this sequence converges and determine the limit.
2. What happens with $a_0=a_1=1$, concerning convergence?

Solution (Proving convergence of a recursively defined sequence)

Solution sub-exercise 1:

There is

$\begin{array}{cc}{a}_n-a_{n-1}& =\frac{1}{3}{a}_{n-1}+\frac{2}{3}{a}_{n-2}-a_{n-1}\\ & =\frac{2}{3}{a}_{n-2}-\frac{2}{3}{a}_{n-1}\\ & =\frac{2}{3}(a_{n-2}-a_{n-1})\\ & =-\frac{2}{3}(a_{n-1}-a_{n-2})\end{array}$

Applying this step $n-1$ times, we get

$\begin{array}{cc}{a}_n-a_{n-1}& =-\frac{2}{3}(a_{n-1}-a_{n-2})\\ & ={(-\frac{2}{3})}^2(a_{n-2}-a_{n-3})\\ & =\dots \\ & ={(-\frac{2}{3})}^{n-1}\underset{=2}{\underbrace{(a_1-a_0)}}\\ & =2\cdot {(-\frac{2}{3})}^{n-1}\end{array}$

We use telescoping sums

$\sum _{k=1}^n(a_k-a_{k-1})=a_n-\underset{=0}{\underbrace{a_0}}=a_n$

and obtain

$\begin{array}{cc}{a}_n& =\sum _{k=1}^n(a_k-a_{k-1})\\ & =2\cdot \sum _{k=1}^n{(-\frac{2}{3})}^{k-1}\\ & =2\cdot \sum _{k=0}^{n-1}{(-\frac{2}{3})}^k\\ & =2\cdot \frac{1-(-\frac{2}{3}{)}^n}{1-(-\frac{2}{3})}\\ & =2\cdot \frac{1-(-\frac{2}{3}{)}^n}{\frac{5}{3}}\\ & =2\cdot \frac{3}{5}(1-(-\frac{2}{3}{)}^n)\\ & =\frac{6}{5}\cdot (1-(-\frac{2}{3}{)}^n)\end{array}$

So we have an explicit form. With $\underset{n\to \mathrm{\infty }}{lim}(-\frac{2}{3}{)}^n=0$ and the limit theorems, we get the limit

$\underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{6}{5}(1-(-\frac{2}{3}{)}^n)=\frac{6}{5}(1-0)=\frac{6}{5}$

Solution sub-exercise 2:

For $a_0=1$ and $a_1=1$ we can plug in

$a_2=\frac{1}{3}\cdot 1+\frac{2}{3}\cdot 1=1$

The next plugging-ins would exactly look the same, and we always get 1. So we assert that the sequence is constantly 1 and therefore converges to 1. We prove this assertion by induction on the pair of consecutive terms:

Theorem whose validity shall be proven for $n\in {\mathbb{N}}_0$ bewiesen werden soll:

$a_n=a_{n+1}=1$

1. Base case:

We start with $n=0$. The initial values give $a_0=a_1=1$.

2. Inductive step:

2a. Inductive hypothesis:

$a_n=a_{n+1}=1$

2b. Induction theorem:

$a_{n+1}=a_{n+2}=1$

2c. Proof of induction step:

$a_{n+2}=\frac{1}{3}\cdot a_{n+1}+\frac{2}{3}{a}_n\stackrel{\text{IH}}{=}\frac{1}{3}\cdot 1+\frac{2}{3}\cdot 1=1$ Together with $a_{n+1}=1$, this proves the induction claim.

Hence, the sequence is constantly 1 and its limit is $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}1=1$.

Application: computing roots by Heron's method

The first sequence elements when computing $\sqrt{2}$ starting at $x_0=2$, $x_0=4$ or $x_0=8$.

The sequence elements of Heron's method for $\sqrt{2}$ when starting at $a_0=\frac{1}{4}$.

We will now come to an application, where convergence of a recursively defined sequence can be shown using the monotonicity criterion: Heron's method can be used to compute roots of positive integers, positive rational numbers or even positive real numbers $a$. It recursively constructs a sequence of approximations for the root $\sqrt{a}$. From the first iterate $x_1$ onwards, these approximations get increasingly better, as shown by the following theorem:

Theorem (Heron's method)

Let $a$ be a real number with $a>0$. Let the sequence $(x_n{)}_{n\in {\mathbb{N}}_0}$ be recursively defined by

$\begin{array}{cc}{x}_0& =a\\ x_{n+1}& =\frac{1}{2}(x_n+\frac{a}{x_n})\end{array}$

Then, $(x_n{)}_{n\in {\mathbb{N}}_0}$ converges to $\sqrt{a}$.

Sequence elements of $(x_n{)}_{n\in \mathbb{N}}$ yield an increasingly precise approximation for $\sqrt{a}$, where we can get to an arbitrary precision $ϵ>0$ . One can, for instance, use a computer to approximate $\sqrt{a}$ up to an arbitrary precision, which is done by performing a sufficiently large number of recursion steps.

Proof (Heron's method)

In order to show that $(x_n{)}_{n\in \mathbb{N}}$ really converges to $\sqrt{a}$ , we will show that:

1. $(x_n{)}_{n\in \mathbb{N}}$ converges by the monotonicity criterion.
2. The limit of $(x_n{)}_{n\in \mathbb{N}}$ is $\sqrt{a}$ .

We can not leave out the first step, as in step 2, we use that the sequence converges. A direct proof of convergence using the Epsilon-criterion (in one step) would be way more complicated.

Proof step: The sequence $(x_n{)}_{n\in \mathbb{N}}$ converges.

Since $x_0=a>0$, induction gives $x_n>0$ for all $n\ge 0$, so the recurrence is well-defined. We prove that the tail $(x_n{)}_{n\in \mathbb{N}}$, starting at $x_1$, is monotonically decreasing and bounded from below.

Proof step: The sequence $(x_n{)}_{n\in \mathbb{N}}$ is bounded from below.

A bound is established using the inequality between arithmetic and geometric means: For $a,b\ge 0$ there is $\frac{a+b}{2}\ge \sqrt{ab}$. Hence,

$\begin{array}{cc}{x}_{n+1}& =\frac{1}{2}(x_n+\frac{a}{x_n})\\ & =\frac{x_n+\frac{a}{x_n}}{2}\\ & \downarrow \frac{a+b}{2}\ge \sqrt{ab}\\ & \ge \sqrt{x_n\cdot \frac{a}{x_n}}\\ & =\sqrt{a}.\end{array}$

So $x_n\ge \sqrt{a}$ for all $n\ge 1$. The full sequence is bounded from below by $min\{x_0,\sqrt{a}\}$.

Proof step: The tail $(x_n{)}_{n\in \mathbb{N}}$, starting at $x_1$, is monotonically decreasing.

This means, we need to show $x_{n+1}\le x_n$ for $n\ge 1$, or equivalently, $x_n-x_{n+1}\ge 0$:

$\begin{array}{cc}{x}_n-x_{n+1}& =x_n-\frac{1}{2}(x_n+\frac{a}{x_n})\\ & =\frac{1}{2}(x_n-\frac{a}{x_n})\\ & =\frac{1}{2x_n}(x_n^2-a)\\ & \downarrow x_n\ge \sqrt{a}>0\Rightarrow \frac{1}{2x_n}>0\\ & =\underset{\ge 0}{\underbrace{\frac{1}{2x_n}}}(x_n^2-a)\\ & \downarrow x_n\ge \sqrt{a}\Rightarrow x_n^2\ge a\Rightarrow x_n^2-a\ge 0\\ & =\underset{\ge 0}{\underbrace{\frac{1}{2x_n}}}\underset{\ge 0}{\underbrace{(x_n^2-a)}}\\ & \ge 0.\end{array}$

So the tail $(x_n{)}_{n\in \mathbb{N}}$, starting at $x_1$, is bounded from below and monotonically decreasing. Hence, it converges by means of the monotonicity criterion. Adding the initial term $x_0$ does not change convergence.

Proof step: The limit of $(x_n{)}_{n\in \mathbb{N}}$ is $\sqrt{a}$.

We will make use of a trick, here: As $(x_n{)}_{n\in \mathbb{N}}$ converges, there must be a limit $x$ , i.e. a real number with $x=\underset{n\to \mathrm{\infty }}{lim}{x}_n$. Then, there is also $\underset{n\to \mathrm{\infty }}{lim}{x}_{n+1}=x$, since $(x_{n+1}{)}_{n\in \mathbb{N}}$ is a subsequence of $(x_n{)}_{n\in \mathbb{N}}$ . The bound $x_n\ge \sqrt{a}$ for $n\ge 1$ implies $x\ge \sqrt{a}>0$, so the quotient rule applies. The recursion relation and the limit theorems yield:

$\begin{array}{cc}x& =\underset{n\to \mathrm{\infty }}{lim}{x}_{n+1}\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{2}(x_n+\frac{a}{x_n})\\ & \downarrow \text{limit theorems}\\ & =\frac{1}{2}(\underset{n\to \mathrm{\infty }}{lim}{x}_n+\frac{a}{\underset{n\to \mathrm{\infty }}{lim}{x}_n})\\ & =\frac{1}{2}(x+\frac{a}{x})\end{array}$

This can be resolved for $x$ and we get the limit:

$\begin{array}{cc}x=\frac{1}{2}(x+\frac{a}{x})& ⟺x=\frac{x^2+a}{2x}\\ & ⟺2x^2=x^2+a\\ & ⟺x^2=a\\ & ⟺x=\pm \sqrt{a}\end{array}$

At this point, there are two candidates for the limit. But since $x_n\ge 0$ , we also have $x\ge 0$ so the convergence must be towards the positive limit candidate

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

Hint

The initial value $x_0$ can be chosen to be any (strictly) positive real number. The proof of convergence works exactly the same way, since $x_1\ge \sqrt{a}$ for every such initial value.

Exercise

Exercise (Recursion for the third root)

Let $x_0\in ℝ$ with $x_0^3>a\in {ℝ}^{+}$. Show that the sequence defined recursively for $n\in {\mathbb{N}}_0$ via

$x_{n+1}=x_n-\frac{x_n^3-a}{3x_n^2}=\frac{1}{3}\cdot (2x_n+\frac{a}{x_n^2})$

converges to $\sqrt[3]{a}$. Proceed as follows:

1. Use Bernoulli's inequality: For every $x\in ℝ$ with $x^3>a$ there is ${(x-\frac{x^3-a}{3x^2})}^3>a$ and use this to show by induction that $x_n^3>a$ for all $n\in {\mathbb{N}}_0$
2. Use step 1 to show: $(x_n{)}_{n\in {\mathbb{N}}_0}$ is monotonically decreasing and bounded from below.
3. Determine the limit of $(x_n)$.

Solution (Recursion for the third root)

1. First inequality: At first,

   ${(x-\frac{x^3-a}{3x^2})}^3={\left(x(1-\frac{x^3-a}{3x^3})\right)}^3=x^3{(1-\frac{x^3-a}{3x^3})}^3$

   Bernoulli's inequality can be applied to the last factor, as

   $-\frac{x^3-a}{3x^3}>-1⟺-(x^3-a)>-3x^3⟺-x^3+a>-3x^3⟺\underset{>0}{\underbrace{a}}>\underset{<-2a<0}{\underbrace{-2x^3}}$

   Set $t=-(x^3-a)/(3x^3)$. Since $-1/3<t<0$, $(1+t{)}^3-(1+3t)=t^2(3+t)>0$, so Bernoulli's inequality is strict here. Thus, with $x^3>a>0$ we get:

   $x^3{(1-\frac{x^3-a}{3x^3})}^3>x^3(1-3\cdot \frac{x^3-a}{3x^3})=x^3-(x^3-a)=a$

   Second inequality: Induction on $n$:

   Induction basis: $n=0$. $x_0^3>a$ holds by assumption.

   Induction step: $n\to n+1$. $x_{n+1}^3={(x_n-\frac{x_n^3-a}{3x_n^2})}^3>a$ by the first inequality, since there is $x_n^3>a$ by the induction assumption.
2. $(x_n)$ is monotonically decreasing: There is $x_{n+1}-x_n=x_n-\frac{x_n^3-a}{3x_n^2}-x_n=-\frac{\stackrel{>0}{\overbrace{x_n^3-a}}}{\underset{>0}{\underbrace{3x_n^2}}}<0$ $(x_n)$ is bounded from below: Since $x_n^3\ge a$ we directly obtain $x_n\ge \sqrt[3]{a}$
3. As $(x_n)$ is monotonically decreasing and bounded from below, the monotonicity criterion implies its convergence. We denote the limit by $\underset{n\to \mathrm{\infty }}{lim}{x}_n=\underset{n\to \mathrm{\infty }}{lim}{x}_{n+1}=x$. Since $x\ge \sqrt[3]{a}>0$, the quotient rule applies. We plug this into the recursive relation and resolve for $x$: $x-\frac{x^3-a}{3x^2}=x⟺\frac{x^3-a}{3x^2}=0⟺x^3-a=0⟺x^3=a⟺x=\sqrt[3]{a}$

Hint

We can even compute roots of higher order: For any natural number $k\ge 2$ , as well as for real numbers $x_0>0$ and $a>0$ we can recursively construct a sequence $(x_n{)}_{n\in {\mathbb{N}}_0}$ with

$x_{n+1}=\frac{1}{k}((k-1)x_n+\frac{a}{x_n^{k-1}})$

and show that it converges to $\sqrt[k]{a}$ .