Convergent Sequences — Introduction to Analysis

We now use the metric structure to describe the asymptotic behavior of an ordered family of points. To say that a sequence $$(x_n)$$ of points in $$M$$ converges to $$x$$ means that its terms eventually lie as close to $$x$$ as we wish.

Definition 3.1.1. Let $$M$$ be a metric space. A sequence $$(x_n)$$ of points in $$M$$ is said to converge in $$M$$ if there is a point $x\in M$ with the following property: For any $$\varepsilon>0$,$ there exists $n_0\in \mathbb Z^{+}$ such that

\[n\ge n_0\ \Longrightarrow\ d(x_n,x)<\varepsilon.\]

In this case, we also say that $$(x_n)$$ converges to $$x$,$ or that $$x$$ is the limit of $$(x_n)$,$ and we write $x_n\to x$ or $$\lim_{n\to \infty} x_n = x$.$ If $$(x_n)$$ does not converge, it is said to diverge in $$M$.$

As can be seen from the definition, convergence depends only on the tail of a sequence: changing finitely many initial terms does not affect either convergence or its limit.

Definition 3.1.2. Let $$M$$ be a metric space. A sequence $$(x_n)$$ in $$M$$ is said to be bounded if the range of $$(x_n)$$ is bounded.

Note that the boundedness of a subset of a metric space is not relative to a subspace when the subspace just carries the restricted metric. Thus the boundedness of a sequence is also not relative to a subspace. We now look into some important properties of convergent sequences in metric spaces.

Theorem 3.1.3. Let $$(x_n)$$ be a sequence in a metric space $$M$.$

The sequence $$(x_n)$$ converges to $x\in M$ if and only if every open ball centered at $$x$$ contains $$x_n$$ for all but finitely many $$n$.$
For $$x,y\in M$,$ if $$(x_n)$$ converges to $$x$$ and to $$y$,$ then $$x=y$.$
If $$(x_n)$$ converges in $$M$,$ then $$(x_n)$$ is bounded.
If $x\in M$ is a limit point of $$A\subseteq M$,$ then there exists a sequence $$(x_n)$$ in $$A$$ that converges to $$x$.$

Proof. (1) Suppose $(x_n)\to x$ and let $$B_r(x)$$ be an open ball of $$x$.$ Corresponding to this $$r$,$ there exists $n_0\in \mathbb Z^+$ such that $n\ge n_0$ implies $$x_n\in B_r(x)$.$ Conversely, suppose that every open ball of $$x$$ contains $$x_n$$ for all but finitely many $$n$.$ Fix $$r>0$$ and let

\[n_0 > \max {\bigl(\{0\}\cup\{ n\in \mathbb Z^+ : x_n\notin B_r(x)\}\bigr)}.\]

Then $n\ge n_0$ implies $$x_n\in B_r(x)$.$ Thus $$x_n\to x$.$ (2) Given $$\varepsilon>0$,$ there exist $n_{0_x},n_{0_y} \in \mathbb Z^+$ such that

\[n\ge n_{0_x} \Rightarrow d(x_n,x)<\frac \varepsilon 2 \quad \text{and} \quad n\ge n_{0_y} \Rightarrow d(x_n,y)<\frac \varepsilon 2.\]

If $$n_0 = \max {\{n_{0_x}, n_{0_y}\}}$,$ then we have

\[d(x,y) \le d(x_n,x) + d(x_n,y) < \varepsilon.\]

Since $\varepsilon$ was arbitrary, we conclude that $$d(x,y)=0$;$ hence $$x=y$.$ (3) Suppose $$x_n\to x$.$ There exists $n_0\in \mathbb Z^+$ such that $$n>n_0$$ implies $$x_n\in B_1(x)$.$ Put

\[\varepsilon = \max {\{d(x_1,x),\ldots,d(x_{n_0},x)\}} + 1.\]

Then we have $d(x_n,x)<\varepsilon$ for all $$n\in \mathbb Z^+$;$ hence $$(x_n)$$ is bounded. (4) For each $$n\in \mathbb Z^+$,$ there is a point $x_n\in A$ such that $$d(x_n,x)<1/n$.$ Given $$\varepsilon>0$,$ choose $$n_0$$ so that $$1/n_0\le \varepsilon$.$ Then, $n\ge n_0$ implies $$d(x_n,x)<\varepsilon$;$ hence $$x_n\to x$.$ $\square$

Definition 3.1.4. A sequence $$(x_n)$$ in $\mathbb R$ is said to be monotonically increasing if $x_n\le x_{n+1}$ for $$n=1,2,3,\ldots$,$ and is said to be monotonically decreasing if $x_n\ge x_{n+1}$ for $$n=1,2,3,\ldots$.$ The sequence $$(x_n)$$ is said to be monotonic if it is monotonically increasing or monotonically decreasing. We often write $x_n\uparrow x$ if $$(x_n)$$ is monotonically increasing and $$(x_n)$$ converges to $$x$.$ Similarly, we write $x_n\downarrow x$ if $$(x_n)$$ is monotonically decreasing and $$(x_n)$$ converges to $$x$.$

Theorem 3.1.5. Suppose $$(x_n)$$ is a monotonic sequence in $$\mathbb R$.$ Then, $$(x_n)$$ converges in $\mathbb R$ if and only if it is bounded.

Proof. Suppose $x_n\le x_{n+1}$ for $n=1,2,3,\ldots$ and let $\alpha$ be the supremum for $$\{x_n\}_{n\in \mathbb Z^{+}}$.$ Given $$\varepsilon>0$,$ by lemma 1.1.10, there is $n_0\in \mathbb Z^{+}$ such that $$x_{n_0}\in (\alpha-\varepsilon,\alpha]$.$ Then, $n\ge n_0$ implies $$|\alpha-x_n|<\varepsilon$;$ hence $$(x_n) \to \alpha$.$ The converse follows from Theorem 3.1.3(3). In the other case, where $$(x_n)$$ is monotonically decreasing, the proof is analogous. $\square$

For $\mathbb C$ with the usual metric $$d(x,y)=|x-y|$,$ taking limits is compatible with its field operations, as the following theorem shows.

Theorem 3.1.6. Suppose $$(x_n)$$ and $$(y_n)$$ are sequences in $$\mathbb C$.$ If $\lim_{n\to \infty} x_n = x$ and $$\lim_{n\to \infty} y_n = y$,$ then the following hold:

$$\displaystyle\lim_{n\to \infty} (x_n + y_n) = x + y$.$
For any $$c\in \mathbb C$,$ we have $$\displaystyle\lim_{n\to \infty} cx_n = cx$.$
$$\displaystyle\lim_{n\to \infty} x_ny_n = xy$.$
If $x_n\ne 0$ and $$x\ne 0$,$ then $$\displaystyle\lim_{n\to \infty}{\frac{1}{x_n}} = \frac 1 x$.$

Proof. (1) Given $$\varepsilon>0$,$ there exist $n_{0_x},n_{0_y} \in \mathbb Z^+$ such that

\[n\ge n_{0_x} \Rightarrow |x_n-x|<\frac \varepsilon 2 \quad \text{and} \quad n\ge n_{0_y} \Rightarrow |y_n-y|<\frac \varepsilon 2.\]

If $$n_0 = \max {\{n_{0_x},n_{0_y}\}}$,$ then $n\ge n_0$ implies

\[| (x_n + y_n) - (x+y)| \le |x_n-x| + |y_n-y| < \varepsilon.\]

Since $\varepsilon$ was arbitrary, (1) holds. (2) Given $$\varepsilon>0$,$ there exists $n_0\in \mathbb Z^+$ such that

\[n\ge n_0 \ \Longrightarrow \ |x_n-x|<\frac{\varepsilon}{|c|+1}.\]

Then, $n\ge n_0$ implies

\[|cx_n - cx| = |c||x_n-x|\le\frac{|c|}{|c|+1}\varepsilon <\varepsilon.\]

Since $\varepsilon$ was arbitrary, (2) holds. (3) Given $$\varepsilon>0$,$ there exist $n_{0_x},n_{0_y}\in \mathbb Z^+$ such that

\[n\ge n_{0_x} \Rightarrow |x_n-x|<\sqrt \varepsilon\quad \text{and} \quad n\ge n_{0_y} \Rightarrow |y_n-y|<\sqrt \varepsilon.\]

If $$n_0 = \max {\{n_{0_x},n_{0_y}\}}$,$ then $n\ge n_0$ implies

\[|(x_n-x)(y_n-y)| = |x_n-x||y_n-y| < \varepsilon.\]

Thus we have $$\lim_{n\to \infty} (x_n-x)(y_n-y) = 0$.$ It follows that

\[\lim\limits_{n\to \infty} (x_ny_n - xy) = \lim\limits_{n\to \infty} ((x_n-x)(y_n-y) + x(y_n-y) + y(x_n-x))=0.\]

Therefore

\[\lim\limits_{n\to \infty} x_ny_n = \lim\limits_{n\to \infty} ((x_ny_n - xy)+xy) = xy.\]

(4) Since

\[\left|\frac {1}{x_n} - \frac{1}{x}\right| = \left|\frac{x_n-x}{x_nx}\right|,\]

what we have to show is to ensure that the denominators $$x_n$$ stay uniformly away from zero. To do that, we first prove the following lemma, so-called the reverse triangle inequality.

Lemma 3.1.7. In a metric space $$(M,d)$,$ the following inequality holds:

\[d(x,y)\ge |d(x,z) -d(y,z)|.\]

Subproof. If $$d(x,z)\ge d(y,z)$,$ let $$\delta=d(x,z)-d(y,z)$.$ In order that

\[d(x,y) + d(y,z) \ge d(x,z) = d(y,z) + \delta ,\]

$$d(x,y)$$ must be greater than or equal to $$\delta$.$ That is, $$d(x,y)\ge d(x,z)-d(y,z)$.$ If $$d(y,z)\ge d(x,z)$,$ in the same way, we have $$d(x,y)\ge d(y,z)-d(x,z)$.$ Therefore $$d(x,y)\ge |d(x,z) -d(y,z)|$.$ $\blacksquare$

Since $$x\ne 0$,$ we have $$d(x,0)>0$.$ Then, there exists $m\in \mathbb Z^+$ such that $n\ge m$ implies $$d(x_n,x)<\frac 1 2 d(x,0)$.$ By Lemma 3.1.7, we obtain

\[d(x_n,0)\ge d(0,x) - d(x_n,x)>\frac 1 2 d(0,x).\]

whenever $$n\ge m$.$ Equivalently, $$|x_n| >\frac 1 2 |x|$.$ Given $$\varepsilon>0$,$ there is $$n_0>m$$ such that $n\ge n_0$ implies

\[|x_n-x|<\frac 1 2 |x|^2 \varepsilon.\]

Hence, if $$n\ge n_0$,$ then

\[\left|\frac {1}{x_n} - \frac{1}{x}\right| = \left|\frac{x_n-x}{x_nx}\right|\le\frac{2}{|x|^2}|x_n-x|<\varepsilon.\tag*{$\square$}\]

Having established the limit laws in $$\mathbb C$,$ we now take Euclidean space into consideration. Because the Euclidean norm is determined by finitely many coordinates, convergence in $\mathbb R^k$ can be tested one coordinate at a time. The next theorem makes this reduction precise.

Theorem 3.1.8. Suppose $(\mathbf x_n)$ is a sequence in $\mathbb R^k$ and $$\mathbf x_n = (x_{1,n},\ldots,x_{k,n})$.$ Then, $(\mathbf x_n)$ converges to $\mathbf x = (x_1,\ldots,x_k)$ if and only if

\[\lim\limits_{n\to \infty} x_{i,n} = x_i\]

for $$i=1,2,\dots,k$.$

Proof. Suppose $$\mathbf x_n\to \mathbf x$.$ From the definition of the norm in $$\mathbb R^k$,$ we have

\[|x_{i,n}-x_i|\le \|\mathbf x_n - \mathbf x\|,\]

for each $i=1,2,\ldots,k$ and $$n\in \mathbb Z^+$.$ Hence, given $$\varepsilon >0$,$ there exists $n_0\in \mathbb Z^+$ such that $n\ge n_0$ implies $$|x_{i,n}-x_i|<\varepsilon$.$ Therefore $x_{i,n}\to x_i$ for each $$i=1,2,\ldots,k$.$ Conversely, suppose $x_{i,n}\to x_i$ for each $$i=1,2,\ldots,k$.$ Given $$\varepsilon>0$,$ there exists $n_{0_i}\in \mathbb Z^+$ such that $n\ge n_{0_i}$ implies

\[|x_{i,n}-x_i| < \frac {\varepsilon}{\sqrt k}.\]

for each $$i=1,2,\ldots,k$.$ Put $$n_0 = \max {\{n_{0_1},\ldots,n_{0_k}\}}$.$ Then $n\ge n_0$ implies

\[\|\mathbf x_n - \mathbf x\| = \left\{\sum_{i=1}^k |x_{i,n}-x_i|^2\right\}^{\!1/2}<\varepsilon.\]

Since $\varepsilon$ was arbitrary, $$\mathbf x_n\to \mathbf x$.$ $\square$

We consider several limits that will be used repeatedly. The following results concern particular sequences in $$\mathbb R$.$ Their proofs also illustrate how the order properties of $$\mathbb R$,$ the binomial theorem, and comparison arguments can be used to establish convergence.

Lemma 3.1.9. Let $$(x_n)$$ and $$(y_n)$$ be sequences in $$\mathbb R$.$ If $0\le x_n\le y_n$ eventually and $$y_n\to 0$,$ then $$x_n\to 0$.$

Proof. Let $\varepsilon>0$ be given. Then eventually $$y_n<\varepsilon$.$ Since $$x_n\le y_n$,$ we have $0\le x_n <\varepsilon$ eventually. Since $\varepsilon>0$ was arbitrary, $$x_n\to 0$.$ $\square$

Theorem 3.1.10.

If $$x>0$$ then $$\displaystyle \lim_{n\to \infty}{\frac {1}{n^x}} = 0$.$
If $$x>0$$ then $$\displaystyle \lim_{n\to \infty}{\sqrt[n]{x}} = 1$.$
$$\displaystyle \lim_{n\to \infty}{\sqrt[n]{n}} = 1$.$
If $$x>0$$ and $y\in \mathbb R$ then $\displaystyle \lim_{n\to \infty}{\frac {n^y}{(1+x)^n}}=0.$
If $$|x|<1$$ then $$\displaystyle \lim_{n\to \infty}{x^n} = 0$.$

Proof. (1) Let $\varepsilon > 0$ be given. The Archimedean property of $\mathbb R$ guarantees the existence of $n_0\in \mathbb Z^+$ such that $$n_0 > \left(1/ {\varepsilon}\right)^{1/x}$.$ Then, $n\ge n_0$ implies $$1/{n^x}<\varepsilon$.$ Since $\varepsilon>0$ was arbitrary, $$1/{n^x}\to 0$.$ (2) If $$x=1$$ then it is trivial. Suppose $$x>1$$ and let $$\sqrt[n]x = 1+\delta_n$,$ so that $$\delta_n>0$.$ Then we have

\[(1 + \delta_n)^n = x.\]

Since

\[1 + n\delta_n \le (1 + \delta_n)^n\]

by the binomial theorem, it follows that

\[1 + n\delta_n \le x.\]

Therefore

\[\delta_n \le \frac {x-1}{n},\]

and thus $\delta_n\to 0$ by Lemma 3.1.9. This proves $$\lim_{n\to \infty}{\sqrt[n]{x}} = 1$.$ Since $$\sqrt[n]x \sqrt[n]{1/x} = 1$,$ we have

\[\lim_{n\to \infty}{\sqrt[n]{\frac 1 x}} = \lim_{n\to \infty}{\frac{1}{\sqrt[n]x}} = \frac{1}{\lim_{n\to \infty}{\sqrt[n]{x}}} = 1.\]

This shows that (2) holds as well when $$x<1$.$ (3) Let $$\sqrt[n]n = 1 + \delta_n$,$ so that $\delta_n>0$ for $$n\ge 2$.$ Then we have

\[(1+\delta_n)^n = n.\]

If $$n\ge 2$,$ then

\[\frac{n(n-1)}{2}\delta_n^2<(1+\delta_n)^n\]

by the binomial theorem. It follows that

\[\frac{n(n-1)}{2}\delta_n^2< n.\]

Thus we have

\[0<\delta_n < \sqrt{\frac {2}{n-1}}.\]

Hence $$\delta_n \to 0$,$ by Lemma 3.1.9. Consequently, $$\sqrt[n]{n} \to 1$.$ (4) Let $$k$$ be an integer such that $$k>y$$ and $$k>0$.$ If $$n>2k$,$ then, by the binomial theorem,

\[(1+x)^n > \binom n k x^k = \frac{n(n-1)\cdots(n-k+1)}{k!}x^k > \frac{n^k}{2^k k!}x^k.\]

It follows that

\[0< \frac {n^y}{(1+x)^n} < \frac{2^k k!}{x^k n^{k-y}}.\]

Since we have

\[\frac{2^k k!}{x^k n^{k-y}} \to 0\]

by (1), Lemma 3.1.9 now gives

\[\frac {n^y}{(1+x)^n} \to 0.\]

(5) If $$x=0$$ it is trivial. If $$0<|x|<1$,$ then taking $$y=0$$ in (4) shows (5). $\square$

Taken together, these assertions describe several basic relations among powers, roots, and exponential sequences. Assertion (1) says that the reciprocal of every fixed positive power of $$n$$ tends to zero. Assertion (2) says that taking successively higher roots removes the scale of any fixed positive number: regardless of its original size, its $$n$-$ th root approaches $$1$.$ Assertion (3) shows that the same conclusion remains true even when the quantity under the root varies with $$n$.$ Thus, although $$n$$ itself tends to infinity, its growth is not fast enough to produce an $$n$-$ th root bounded away from $$1$.$ Assertions (4) and (5) describe two sides of exponential behavior. In (4), $$n^y$$ is a fixed power of $$n$,$ whereas $$(1+x)^n$$ is an exponential sequence with fixed base greater than $$1$.$ Both $$x$$ and $$y$$ are held fixed as $$n\to\infty$.$ Therefore, no matter how close $$1+x$$ is to $$1$,$ or how large $$y$$ is, the exponential term eventually dominates the power term. Assertion (5) gives the corresponding statement for exponential decay: repeated multiplication by a fixed number whose absolute value is less than $$1$$ forces the resulting sequence to zero.