Real Analysis/Sequences

Definition
Sequences occur frequently in analysis, and they appear in many contexts. While we are all familiar with sequences, it is useful to have a formal definition.


 * Definition A sequence of real numbers is any function a : N→R.

Often sequences such as these are called real sequences, sequences of real numbers or sequences in R to make it clear that the elements of the sequence are real numbers. Analogous definitions can be given for sequences of natural numbers, integers, etc.

Given a sequence (xn), a subsequence, notated as $$(x_{n_j})_{j=1}^\infty$$, is a sequence where (nj) is strictly increasing sequence of natural numbers.

For example, taking nj=2j would the subsequence consisting of every other element of the original sequence, that is (x2, x4, x6, …).

Sequence Notation
However, we usually write an for the image of n under a, rather than a(n). The values an are often called the elements of the sequence. To make a distinction between a sequence and one of its values it is often useful to denote the entire sequence by $$(a_n)_{n=1}^\infty$$, or just (an). Some employ set notation and denote it as {an} When specifying a particular sequence, it may be written in the form (a1, a2, a3, …), when the sequence is infinite, or (a1, a2, …, an) when the sequence is finite. We tend only to discretely write down enough elements is so that the pattern is clear, which is typically 3 times.

Examples of Sequences
(1, 2, 3, 4, …), (1, -2, 3, -4, …), and (1, π, π2, π3, π4, …) are all examples of sequences. Note, however, that there need not be any particular pattern to the elements of the sequence. For example, we may specify an to be the n-th digit of π. Often sequences are defined recursively. That is, to specify some initial values of the sequence, and then to specify how to get the next element of the sequence from the previous elements. For example, consider the sequence x1=1, x2=1, and xn = xn&minus;1 + xn&minus;2 for n &ge; 3. This sequences is known as the Fibonacci sequence, and its first few terms are given by (1, 1, 2, 3, 5, 8, 13, …). Another familiar example of a recursive sequence is Newton's method. With an initial guess x0 for the zero of a function, Newton's method tells you how to construct the next guess. In this way you generate a sequence which (hopefully) converges to the zero of the function.

Operations on Sequences
We can also perform algebraic operations on sequences. In other words, we can add, subtract, multiply, divide sequences. These operations are simply performed element by element, for completeness we give the definitions.


 * Definition Given two sequences (xn) and (yn) and a real number c, we define the following operations:

Classification of Sequences
Some properties of sequence are so important that they are given special names.

Some of these terms are prefixed with strictly because the term increasing is used in some contexts with meaning either that of strictly increasing or of non-decreasing, and similarly decreasing can mean the same as either strictly decreasing, or non-increasing. As a result, these ambiguous terms are usually prefixed with and strictly. We will try adhere to using this unambiguous term.

From here, we will also describe properties of sequences based on boundedness, a word which we will define for sequences below.

Convergence and Limits
A further important property of sequences (arguably the most important property from the perspective of analysis) is the property of convergence. This property can be easily described by extending the epsilon-delta definition. However, because sequences are relative to counting numbers, there exists an additional way to imagine convergence. Both methods are described below.


 * Definition Let (xn) be a sequence of real numbers. The sequence (xn) is said to converge to a real number a.
 * if for all ε>0, there exists N in N such that |xn-a|&lt;ε for all n&ge;N.

If (xn) converges to a then we say a is the limit of (xn) and write


 * $$\lim_{n\to\infty}x_n=a$$

or


 * $$x_n\to a$$ as $$n\to\infty$$.

This is read xn approaches a as n approaches &infin;. If it is clear which variable is playing the role of n then this may be abbreviated to simply xn→a or lim xn=a.

If a sequence converges, then it is called convergent.

It is also useful to extend this concept and allow sequences whose limits are either &infin; or &minus;&infin;


 * Definition We say xn→&infin; as n→&infin; if for every M in R there is a natural number N so that xn&ge; M for all n&ge;N.  We say xn→&minus;&infin; as n→&infin; if for every M in R there is a natural number N so that xn&le; M for all n&ge;N.

Despite this, we do not refer to sequences such as these as convergent. They are instead called divergent.

Although convergence can be proven using the epsilon-delta definition as proof, another method to prove convergences of sequences is through mathematical induction, since sequences are referenced using counting numbers. Through this method, some theorems are easier to prove. However, proof using mathematical induction cannot generalize to real numbers like a proof using epsilon-delta can.

The following theorems will prove that variations of a convergent sequence, expressed either through inductive notation, limit notation, or Cauchy notation, converges to exactly one number. This may seem intuitively clear, but remember that intuition often fails us when it comes to limits. It is also in proper mathematical style to rigorously prove every mathematical notion presented to us.

Theorem (Uniqueness of limits)
A sequence can have at most one limit. In other words: if xn → a and xn → b then a = b.

Proof
Suppose the sequence has two distinct limits, so a&ne;b. Let ε=|a&minus;b|/3.

Certainly ε&gt;0, using the definition of convergence twice we can find natural numbers Na and Nb so that


 * $$|x_n-a|\leq\epsilon$$ for all n &gt; Na.

and


 * $$|x_n-b|\leq\epsilon$$ for all n &gt; Nb.

Taking k=max(Na,Nb) then both of these conditions hold for xk. Hence we deduce that |xk&minus;a|&le;ε and |xk&minus;b|&le;ε. Applying the triangle inequality, we see
 * $$|a-b|=|(x_k-b)-(x_k-a)|\leq 2\epsilon=(2/3)|a-b|,$$

which is a contradiction. Thus, any sequence has at most one limit. $$\Box$$

Theorem (Convergent Sequences Bounded)
If the subsequence $$(x_n)_{n=1}^{\infty}$$ is a convergent sequence, then it is bounded.

Proof
Let $$a=\lim_{n\to\infty}x_n$$, and let ε = 1.

From the definition of convergence there exists a natural number N such that


 * $$|x_n-a|\leq 1$$ for all n &ge; N.

The sequence $$(x_n)_{n=N+1}^{\infty}$$ is bounded above by a+1 and below by a&minus;1. Let M = max(|x1|,|x2|,|x3|,…,|xN|, |a|+1). It follows that &minus;M &le; xn &le; M for all n in N. Hence the sequence is bounded. $$\Box$$

Theorem (Boundedness of Cauchy Sequences)
If $$(x_n)_{n=1}^{\infty}$$ is a Cauchy sequence, then it is bounded.

Proof
Let (xn) be a Cauchy sequence. By the definition of a Cauchy sequence, there is a natural number N such that |xn&minus;xm|&lt;1 for all n,m &gt; N. In particular, |xN+1&minus;xm|&lt;1 for all m &gt; N. It follows by the reverse triangle inequality that |xm| &lt; |xN+1| + 1. If we take M=max(|x1|, |x2|, …, |xN|, |xN+1| + 1), then |xn| &le; M for all n in N. $$\Box$$

The following theorem tells us that algebraic operations on sequences commute with the taking limits. This simple theorem is a useful tool in computing limits.

Properties of Sequences
Given our new definition of convergence, it should be essential that we can use the values we get from them algebraically and whether or not we can apply algebraic intuition in regards to converging sequences as well.

Algebraic Operations
If (xn) and (yn) are convergent sequences and a &isin; R, the following properties hold:
 * 1) $$\lim_{n\to\infty}(x_n+y_n)=\lim_{n\to\infty}(x_n)+\lim_{n\to\infty}(y_n)$$.
 * 2) $$\lim_{n\to\infty}(x_ny_n)=\left(\lim_{n\to\infty}x_n\right)\left(\lim_{n\to\infty}y_n\right)$$.
 * 3) $$\lim_{n\to\infty}(ax_n)=a\lim_{n\to\infty}(x_n)$$.
 * 4) $$\lim_{n\to\infty}\left(\frac{x_n}{y_n}\right)=\frac{\displaystyle \lim_{n\to\infty}x_n}{\displaystyle \lim_{n\to\infty}y_n}$$ (assuming yn &ne; 0 for all n in N and lim y_n &ne; 0).
 * 5) If xn &le; yn for every n in N, then $$\lim_{n\to\infty}x_n \leq \lim_{n\to\infty}y_n$$.

Proof
1. Let x=lim xn and y=lim yn. We need to show that for any ε&gt;0 there is natural number N so that if n&ge; N, then |(xn + yn) &minus; (x + y)|&le;ε. Given any ε&gt;0 we have ε/3&gt;0 so from the definition of convergence there is a natural number Nx so that |xn&minus;x|&le;ε/3 for all n&gt;Nx, similarly we can choose Ny |yn&minus;y|&le;ε/3 for all n&gt;Ny.

Let N=max(Nx ,Ny). If n&gt;N, then by the triangle inequality we have
 * $$|(x_n+y_n)-(x+y)|=|(x_n-x)+(y_n-y)| \leq \epsilon /3 + \epsilon /3<\epsilon,$$

which is what we needed to show. $$\Box$$

2. Let x=lim xn and y=lim yn. Since these sequences are convergent they are bounded. Let Mx be a bound for (xn) and let My be a bound for (yn). By increasing these quantities of necessary we may also assume Mx &gt; x and My &gt; y. Given ε&gt;0, there exists some Nx and Ny such that
 * $$|x_n - x| < {\epsilon \over 2M_y}$$ for n &gt; Nx and
 * $$|y_n - y| < {\epsilon \over 2M_x}$$ for n &gt; Ny.

Then for every n &gt; max(Nx, Ny),

$$\Box$$
 * $$|x_n y_n - xy|\,$$
 * $$= |(x_n - x)y_n + x(y_n - y)|\,$$
 * $$\leq |x_n - x|M_y + M_x |y_n - y|$$
 * $$\leq \frac{\epsilon}{2} + \frac{\epsilon}{2}\leq \epsilon.$$
 * }
 * $$\leq |x_n - x|M_y + M_x |y_n - y|$$
 * $$\leq \frac{\epsilon}{2} + \frac{\epsilon}{2}\leq \epsilon.$$
 * }
 * $$\leq \frac{\epsilon}{2} + \frac{\epsilon}{2}\leq \epsilon.$$
 * }

3. Let yn = a for all n in N. The statement now follows from 2. $$\Box$$

4. We can reduce this to showing that lim (1/yn) exists and equals 1/(lim yn). Then it follows by 2 that we have:
 * $$\lim_{n\to\infty}\left(\frac{x_n}{y_n}\right)= \left( \lim_{n \to \infty} {1 \over y_n} \right) \left( \lim_{n \to \infty} {x_n} \right)=\frac{\displaystyle \lim_{n\to\infty} x_n}{\displaystyle \lim_{n\to\infty} y_n}.$$

Let y=lim yn. By the exercises, since y and yn are not 0, we can find δ &gt; 0 so that |y_n| &gt; δ and |y| &gt; δ. It follows that 1/|yny|<1/δ2. Given ε &gt; 0 choose n in N so that |yn &minus; y| &lt; δ2ε. We have
 * $$\left| {1 \over y_n} - {1 \over y} \right| = |y - y_n|{1 \over |y_n y|} \le \frac{|y - y_n|}{\delta^2} < \epsilon$$.

Hence, $$\lim_{n \to \infty} {1 \over y_n} = {1 \over y}.$$ $$\Box$$

5. We first can reduce to the case when one sequence is identically 0. To see this let zn = xn &minus; yn. Then zn &lt; 0 for all n in N. Let z = lim zn. Suppose that z &gt; 0 then we can then find a natural number N so that
 * $$|z - z_N| < z\,$$.

Since zN &le; 0 &lt; z, the absolute value equals z &minus; zN. Subtracting z we find that &minus;zN &lt; 0. Hence zN is positive. Contradiction. Therefore we must have that z &le; 0. Which means that by 1 we get:
 * $$\lim_{n \to \infty} x_n - \lim_{n \to \infty} y_n = \lim_{n \to \infty} z_n = z \le 0.$$

Therefore lim xn &le; lim yn $$\square$$

Theorem (Squeeze/Sandwich Limit Theorem)
This is the important squeeze theorem that is a cornerstone of limits. Since converging sequences can also be thought of through limit notions and notations, it should also be wise if this important theorem applies to converging sequences as well.

Given sequences (xn), (yn), and (wn), if (xn) and (yn) converge to a and xn &le; wn &le; yn, then wn converges to a.

Proof
Fix ε &gt; 0. We need to find an N such that |wn &minus; a| &lt; ε if n &gt; N. Since (xn) → a and (yn) → a the definition of convergence ensures that there exists integers Nx and Ny so that |xn &minus; a| &lt; ε for n &gt; Nx and |yn &minus; a| &lt; ε for n &gt; Ny.

Let N=max(Nx, Ny). Then, for all n &gt; N we have &minus;ε &lt; xn &minus; a and yn &minus; a &lt; ε. Since xn &lt; wn &lt; yn, it follows that xn &minus; a &lt; wn &minus; a &lt; yn &minus; a.

Thus if n &ge; N, then &minus;ε &lt; xn &minus; a &lt; wn &minus; a &lt; yn &minus; a &lt; ε. In other words, |wn &minus; a| < ε.$$\Box$$

Completeness
The following results are closely related to the completeness of the real numbers.

Theorem (Convergence of Monotone sequences)
Any monotone, bounded sequence converges. If the sequence is non-decreasing, then the sequence converges to the least upper bound of the elements of the sequence. If the sequence is non-increasing, then the sequence converges to the greatest lower bound of the elements of the sequence

Proof
Let (xn) be any monotone sequence that is bounded by a real number M. Without loss of generality, assume (xn) is non-decreasing. Since (xn) is bounded above, it has a least upper bound by the least upper bound axiom. Let x = sup {xn | n &isin; N}. We will now show that (xn) → x.

Fix ε &gt; 0. As was shown in the exercises, if s = sup(A), then for any ε &gt; 0 there is an element a in A  so that  s &minus; ε &lt; a &lt; s. Hence, it follows that there exists an N in N so that x &minus; ε &lt; xN &lt; x.

For any n &gt; N, since xn is non-decreasing, we have that
 * $$x- \epsilon < x_N \leq x_n < x$$.

Thus |x &minus; xn| &lt; ε and by the definition of convergence, (xn) converges to x. $$\Box$$

Theorem (Nested intervals property)
If there exists a sequence of closed intervals In = [an, bn] = {x | an &le; x &le; bn} such that In+1 &sube; In for all n, then &cap;In is nonempty.

Proof
Since In+1 &sube; In it follows that an &le; an+1 and bn+1 &le; bn.

Since (an) and (bn) are monotonic sequences they converge by the previous theorem. Furthermore, since an &lt; bn for all n, it follows that lim an &le; lim bn.

By the monotonicity of (an) and (bn) we have for every n
 * $$a_n \leq \lim_{n\to\infty}a_n \leq \lim_{n\to\infty}b_n \leq b_n.$$

Therefore lim an &isin; [an, bn] for every n, which implies that
 * $$\lim_{n\to\infty}a_n \in \bigcap _{n=1}^{\infty} I_n.$$

Thus the intersection is nonempty. $$\Box$$

Theorem (Bolzano&mdash;Weierstrass)
Every bounded sequence of real numbers contains a convergent subsequence.

Proof
Let (xn) be a sequence of real numbers bounded by a real number M, that is |xn| &lt; M for all n. We define the set A by A = {r | |r| &le; M and r &lt; xn for infinitely many n}. We note that A is non-empty since it contains &minus;M and A is bounded above by M. Let x = sup A.

We claim that, for any ε &gt; 0, there must be infinitely many points of xn in the interval (x &minus; ε, x + ε). Suppose not and fix an ε &gt; 0 so that there are only finitely many values of xn in the interval (x &minus; ε, x + ε). Either x &le; xn for infinitely many n or x &le; xn for at most only finitely many n (possibly no n at all). Suppose x&lt; xn for infinitely many n. Clearly in this case x &ne; M. If necessary restrict ε so that x + ε &le; M. Set r = x + ε/2 we have that r &lt; xn for infinitely many n because there are only finitely many xn in the set [x,r] and x must be less than infinitely many xn, furthermore |r| &lt; M. Thus r is in A, which contradicts that x is an upper bound for A. Now suppose x&lt; xn for at most finitely many n. Set y = x &minus; ε/2. Then there are at most only finitely man n so that xn &ge; y. Thus, if r &lt; xn for infinitely many n, we have that r &le; y. This means that y is an upper bound for A that is less than x, contradicting that x wast the least upper bound of A. In either case we arrive at a contradiction, thus we must have that for any ε &gt; 0, there must be infinitely many points of xn in the interval (x &minus; ε, x + ε). Now we show there is a subsequence that converges to x. We define the subsequence inductively, choose any xn 1 from the interval (x &minus; 1, x + 1). Assuming we have chosen xn 1, …, xn k&minus;1 , choose xn k to be an element in the interval (x &minus; 1/k, x + 1/k) so that nk&notin;{n1, …, nk&minus;1}, this is possible as there are infinitely many elements of (xn) in the interval. Notice that for this choice of xn k we have that |x &minus; xn k |<1/k. Hence for any ε>0, if we take any k &gt; 1/ε, then |xn k -x| &lt; ε. That is the subsequence (xn k ) → x. $$\Box$$

Theorem (Cauchy criterion)
A sequence converges if and only if it is Cauchy. Although this seems like a weaker property than convergence, it is actually equivalent, as the following theorem shows:

Proof
First we show that if (xn) → x then xN is Cauchy. Now suppose that for a given ε &gt; 0 we wish to find an N so that |xn &minus; xm| &lt; ε for all n, m &gt; N. We will choose N so that for all n &ge; N we have that |xn &minus; x| &lt; ε/2. By the triangle inequality, for any n, m &gt; N we have:


 * $$|x_n - x_m| \leq |x_n - x| + |x_m - x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$.

Thus (xn) is a Cauchy sequence.

Now we show that if (xn) is a Cauchy sequence, then it converges to some x. Let (xn) be a Cauchy sequence, and let ε &gt; 0. By the definition of a Cauchy Sequence, there exits a natural number L so that |xn &minus; xm| &lt; ε/2 whenever n, m &gt; L. Since (xn) is a Cauchy sequence it is bounded. By the Bolzano&mdash;Weierstrass theorem, it has a convergent subsequence (xn k ) that converges to some point x. Now we will show that the whole sequence converges to x

Because (xn k ) converges, we can choose a natural number M so that if nk &gt; M, then |xn k &minus; x| &lt; ε/2. Let N = max(L, M), and fix any nk &gt; N. For n &gt; N we have that


 * $$|x_n - x| \leq |x_n - x_{n_k}| + |x_{n_k} - x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$.

Thus by definition of convergence (xn) → x. $$\Box$$

These theorems all describe different aspects of the completeness of the real numbers. The reader will notice that the least upper bound property was used heavily in this section, and it is the axiom that separates the real numbers from the rational numbers. While these theorems would be false for the rational numbers, not all of them can substitute for the least upper bound property. The Cauchy criterion and the nested intervals property are not strong enough to imply the least upper bound property without additional assumptions, while the Convergence of Monotone sequences theorem and the Bolzano&mdash;Weierstrass property do imply the least upper bound property.

Limit superior and limit inferior
Limits turn out to be a very useful tool in analysis, their primary draw back is that they may not always exist. Occasionally it is useful to have some notion of limit that makes sense for any sequence. To this end we introduce the limit superior (often just called the "lim sup") and the limit inferor (often called the "lim inf").

Definition For a sequence (xn) we define the limit superior, denoted lim&thinsp;sup by:
 * $$\limsup_{n\to\infty} x_n=\inf_{k}\sup_{n\geq k} x_n.$$

Similarly we define the limit inferior, denoted by lim&thinsp;inf by:
 * $$\liminf_{n\to\infty} x_n=\sup_{k}\inf_{n\geq k} x_n.$$

If (xn) is not bounded above, we say that lim&thinsp;sup xn = &infin;. If (xn) is not bounded we say that lim&thinsp;inf xn = &minus;&infin;.

Notice that for bounded sequences the lim&thinsp;sup and the lim&thinsp;inf always exist. As we know general bounded sequence the limit doesn't always exist. But in the case when the lim&thinsp;sup and lim&thinsp;inf are equal, life is nicer as the next theorem shows.

Theorem (Limit Superior and Inferior)
Let (xn) be a bounded sequence. Then (xn) → x if and only if lim&thinsp;sup xn = x = lim&thinsp;inf xn.

Proof
First suppose (xn) → x. Fix an ε &gt; 0 choose a natural number N so that x &minus; ε &lt; xn &lt; x + ε for any n &gt; N. Hence for any k &gt; N we have that
 * $$x-\epsilon\leq \sup_{n>k}x_n\leq x+\epsilon$$

and hence x &minus; ε &lt; lim&thinsp;sup xn &lt; x + ε. Since ε was arbitrary, this can only happen if lim&thinsp;sup xn = x. A similar argument shows that lim&thinsp;inf xn = x.

Now suppose lim&thinsp;inf xn = x = lim&thinsp;sup xn, and we wish to show that lim xn = x.

First recall that the x=lim&thinsp;sup xn is defined as:
 * $$x=\inf_k\sup_{n>k} x_n$$

Given an ε &gt; 0, since we can get arbitrarily close to the infimum, we can choose we will choose Nls so that
 * $$\sup_{n>N_{ls}} x_n - x < \epsilon$$

Similarly recall that the x=lim&thinsp;inf xn is defined as:
 * $$x=\sup_k\inf_{n>k} x_n$$

Since we can get arbitrarily close to the supremum, we can choose we will choose Nli so that
 * $$x-\sup_{n>N_{li}} x_n< \epsilon$$

Let N = max(Nls, Nli). Now if n &gt; N, then
 * $$\inf_{n> N_{li}}\leq \inf_{n\geq N}\leq x \leq \sup_{n\geq N}\leq \sup_{n> N_{ls}}$$

Hence for any n &gt; N
 * $$x-\sup_{n> N_{ls}} x_n \leq x-x_n\leq x-\inf_{n>N_{li}} x_n.$$

By our choice of Nls and Nli this implies for any n &gt; N


 * $$-\epsilon \leq x-x_n\leq \epsilon.\quad\Box$$