Math 521 Sequences in Metric spaces

The real number line $\R$ with the usual distance function $d(x,y) = |x-y|$ is a metric space, so all definitions and theorems about convergent sequences in metric spaces apply to sequences of real numbers. To summarize, at this point we have the following criteria for showing that a sequence converges:

The definition: a sequence $x_n\in\R$ converges to $a\in\R$ if for every $\epsilon\gt0$ there is an $N_\epsilon\in\N$ such that for all $n\ge N_\epsilon$ one has $|x_n-a|\lt \epsilon$.
Monotonicity: a monotone and bounded sequence always converges (see here for the proof.)
Cauchy property: if $x_n$ is a Cauchy sequence then it converges; conversely, every convergent sequence is a Cauchy sequence.

The calculus limit properties. The limit laws that are familiar from calculus all are true. For example, if $x_i$ and $y_i$ are convergent sequences of real numbers, then their sum, difference, and product also are convergent sequences, and one has \[ \lim_{i\to\infty} x_i\pm y_i = \lim_{i\to\infty} x_i \pm \lim_{i\to\infty} y_i \quad \text{ and } \quad \lim_{i\to\infty} \bigl(x_iy_i\bigr) = \left(\lim_{i\to\infty} x_i \right)\left( \lim_{i\to\infty} y_i\right). \] If $y_i\neq0$ for all $i\in\N$ and if $\lim y_i\neq 0$, then $x_i/y_i$ is well defined, and converges with \[ \lim_{i\to\infty}\frac{x_i}{y_i} = \frac{\lim_{i\to\infty} x_i}{\lim_{i\to\infty} y_i}\;. \]

Lim sup and lim inf of a sequence. If $x_n\in\R$ is a bounded sequence of real numbers then we can define the sequences \begin{align} m_n = \inf \{x_n, x_{n+1}\, \dots\} = \inf_{k\geq n} x_k,\qquad M_n = \sup \{x_n, x_{n+1}\, \dots\} = \sup_{k\geq n} x_k. \end{align} Then sequence $m_n$ is nondecreasing, and the sequence $M_n$ is nonincreasing (why?). Both sequences are bounded, and therefore they converge. By definition \[ \liminf_{n\to\infty} x_n = \lim_{n\to\infty}\left(\inf_{k\geq n} x_k\right),\qquad \limsup_{n\to\infty} x_n = \lim_{n\to\infty}\left(\sup_{k\geq n} x_k\right). \]

Theorem. If $x_n\in \R$ is a bounded sequence of real numbers, then \[ \liminf_{n\to\infty} x_n \leq \limsup_{n\to\infty} x_n\,. \] Moreover, $\liminf x_n = \limsup x_n$ holds if an only if the sequence $x_n$ converges. When this happens we have \[ \liminf x_n = \limsup x_n = \lim x_n\,. \]

The fat sandwich theorem. Let $a_n$ and $b_n$ be two convergent sequences, and let $x_n$ be a third sequence which satisfies $a_n \leq x_n \leq b_n$ for all $n\in\N$. Then \[ \lim a_n \leq \liminf x_n \leq \limsup x_n \leq \lim b_n\,. \] In particular, if the two sequences $a_n$ and $b_n$ have the same limit, then \[ \lim a_n = \lim x_n = \lim b_n\,. \]

Theorem. Let $x_n$ be a bounded sequence of real numbers, and suppose that $x_{n_k}$ is a convergent subsequence of $x_n$. Then \[ \liminf_{n\to\infty} x_n \leq \lim_{k\to\infty} x_{n_k} \leq \limsup_{n\to\infty} x_n\,. \]

Warning: excessive use of $\infty$ can lead to bad algebra. The symbol $\infty$ for “infinity” is very useful, but careless use can quickly lead to contradictions such as this: since $x-x=0$ for any quantity $x$, and since the sum of two infinities clearly must again be infinite, we have \[ \infty + \infty = \infty \implies \infty + \not\infty -\not\infty = \not\infty-\not\infty \implies \infty = 0\,?! \] The simplest way to avoid such contradictions is to avoid doing arithmetic or algebra with the symbol $\infty$ completely, and to be clear about what we mean when we do use the symbol. The following definitions explain what we mean when we say that $x_n\to\infty$, or $\limsup x_n = \infty$.

Definition. If $x_n$ is a sequence of real numbers, then we say $x_n\to\infty$, or $\lim_{n\to\infty} x_n=\infty$, if for every $A\in\R$ there is an $N_A\in\N$ such that for all $n\ge N_A$ one has $x_n\gt A$.

We say that $\lim_{n\to\infty} x_n = -\infty$ if for every $A\in\R$ there is an $N_A\in\R$ such that for all $n\geq N_A$ we have $x_n\lt A$.

Once we adopt this last definition the least upper bound axiom becomes simpler, namely, it says that every nonempty set $E$ of real numbers has a least upper bound $\sup E$, i.e. we no longer need to assume that $E$ has an upper bound, because if it doesn’t, then by definition $\sup E=\infty$. The danger of this definition is that we could be doing a proof in which we set $m=\sup E$ for some $E\subset \R$, forget that $m$ could be infinite, and then do a computation with $m$ (e.g. $m-m=0$). The end result could then be a very confusing contradiction.

A sequence $x_n$ is called a geometric sequence if $x_n\neq0$ for all $n\in\N$, and if the ratio $r=\frac{x_{n+1}}{x_n}$ is the same for all $n$. If $x_n$ is a geometric sequence, then \[ x_n = x_0 r^n . \]

Proof. Assume $0\lt r\lt 1$. The sequence $x^n$ is monotone, because $0\lt r\lt 1$ implies $r^{n+1}\lt r^n$. Therefore $a=\lim_{n\to\infty} r^n$ exists. We have \[ ar = r\lim_{n\to\infty} r^{n} = \lim_{n\to\infty} r^{n+1}. \] The sequence $r^{n+1}$ is a subsequence of $r^n$, so it also converges and has the same limit. We conclude $ar=a$. Since $r\neq 1$ this implies $a=0$.

If $|r|\lt 1$ then we have just shown that $|r|^n\to0$. From $-|r|^n \leq x^n \leq |r|^n$ it then follows that $r^n\to0$. ////

Comparison with geometric sequences. The previous theorem about convergence of $x^n$ can be applied to other sequences by comparing them with geometric sequences. Here is a theorem that enables such a comparison: if $x_n$ is a sequence of real numbers for which \[ \limsup_{n\to\infty} \left|\frac{x_{n+1}}{x_n}\right| = a \] holds, then for every $r\gt a$ there is a $C\in\R$ such that for all $n\in\N$ one has \[ |x_n| \leq C r^n. \]

In particular, if $a = \limsup |x_{n+1}/x_n| \lt 1$ then we can choose $r$ so that $a\lt r\lt 1$, and conclude that $|x_n|\leq Cr^n$ for all $n\in\N$. Since $r\lt 1$ we know that $r^n\to0$, and therefore we get $x_n\to 0$.

Often the sequence of ratios $|x_{n+1}/x_n|$ converges, in which case the lim sup is the same as the limit, i.e. \[ \text{if }\lim \left|\frac{x_{n+1}}{x_n}\right|\text{ exists, then } \limsup \left|\frac{x_{n+1}}{x_n}\right| = \lim \left|\frac{x_{n+1}}{x_n}\right|. \]

Proof. To prove the theorem, we unwrap the hypothesis. The definition of “limsup” tells us that \[ \lim_{n\to\infty}\left(\sup_{k\geq n} \left|\frac{x_{k+1}}{x_k}\right|\right) = a, \] so that for any $\epsilon\gt0$ there is an $N_\epsilon\in\N$ such that for all $n\geq N_\epsilon$ we have \[ \sup_{k\geq n} \left|\frac{x_{k+1}}{x_k}\right| \lt a+\epsilon. \] If we choose $\epsilon = r-a$ then $a+\epsilon = r$ and we know that for all $n\geq N_\epsilon$ we have \[ \left|\frac{x_{n+1}}{x_n}\right| \leq \sup_{k\geq n} \left|\frac{x_{k+1}}{x_k}\right| \lt r. \] Now consider the numbers $y_n = |x_n|/r^n$. We have for all $n\geq N_\epsilon$ \[ \frac{y_{n+1}}{y_n} = \frac{|x_{n+1}|/r^{n+1}}{|x_n|/r^n} = \frac 1r \left|\frac {x_{n+1}}{x_n}\right| \lt 1. \] In other words, for $n\geq N_\epsilon$ we have $y_{n+1} \lt y_n$, i.e. the $y_n$ are decreasing for $n\geq N_\epsilon$. This implies that $y_n\leq y_{N_\epsilon}$ for all $n\geq N_\epsilon$. Now define \[ C = \max \left\{ y_1, y_2, \dots, y_{N_\epsilon}\right\}. \] Then we have $y_n\leq C$ for all $n\in\N$, and thus $|x_n| = r^n y_n \leq Cr^n$, as claimed. ////

Proof. Let $x_n = n^k/a^n$ and compute \[ \lim_{n\to\infty} \frac{x_{n+1}}{x_n} = \lim_{n\to\infty} \left(1+\frac1n\right)^k \frac 1a = \frac 1a \lt 1. \] It then follows that there exist $C\in\R$ and $\frac1a\lt r\lt 1$ such that for all $n\in\N$ \[ x_n \leq C r^n \] holds. Therefore $\limsup_{n\to\infty} x_n \leq \lim_{n\to\infty}Cr^n=0$.

Sequences of real numbers