Real Analysis: Sequences
Convergent and Divergent Sequences
Motivation
One of most important properties of a sequence is whether it eventually approaches a particular value. When all the terms of a sequence after a certain point are within an arbitrary level of closeness to a value, the sequence is said to converge to that value. For example, the sequence $\{\frac{1}{n}\}$ gets closer and closer $0$ as $n$ grows larger, and so we say that $\{\frac{1}{n}\}$ converges to $0.$
Of dual concern is whether a sequence does not converge to a particular value. The sequence $\{n^2\}$ clearly does not get closer to any particular real number and instead grows indefinitely. But what about in the extended real numbers? The sequence $\{n^2\}$ seems to "converge" to infinity. Does this count? Let's clarify with proper definitions of convergence and divergence.
Convergent Sequences
A sequence $\{a_n\}$ converges to a point $p$ in a metric space $X$ if for every positive $\varepsilon \in \mathbb{R}$ there exists an $N \in \mathbb{N}$ such that $d(a_n, p) < \varepsilon$ for all $n > N$. If $\{a_n\}$ converges to some $p$, we can equivalently say that the limit of $\{a_n\}$ as $n$ approaches infinity is $p$, and we can write this as $\lim\limits_{n \rightarrow \infty} a_n = p$, or simply $\lim a_n = p$.
Dependence on Metric Space
The definition of convergence is made in terms of a general metric space $X$, rather than $\mathbb{R}$, because, as shown in the problems below, the convergence of a sequence is in fact dependent on the underlying metric. While this general definition covers the essence of any kind of convergent sequence, determining the convergence a sequence in a particular metric space, such as $\mathbb{R}$ under the standard Euclidean metric, requires using the particular facts about that metric.
Divergent Sequences
A sequence diverges if it does not converge. A sequence that diverges is said to be divergent. If a sequence diverges, then its limit does not exist.
These definitions resolve the initial question of whether a sequence can "converge" to infinity. No, it cannot, because convergence formalizes a "getting closer to" notion. Since infinity is by definition infinitely far away from every real number, each element of the sequence is in fact no closer than the last. However, we can formalize this notion of "convergence to infinity" by proposing an alternate definition.
A sequence of real numbers $\{a_n\}$ tends to infinity (or explodes) if for every real $k \in \mathbb{R}$ there is some $N \in \mathbb{N}$ such that $a_n > k$ for all $n > N$. In this instance, we write $\lim\limits_{n \rightarrow \infty} a_n = \infty$. Plainly, a sequence tends to infinity if for any real number, there is a point at which all the terms in the sequence exceed it. An analogous definition holds for sequences that tend to negative infinity. This definition formalizes a "getting farther away from" notion or an "unending growth" notion, which fits the phenomenon better. While the distance to infinity never diminishes, the distance from $0$ certainly grows!
Note: A sequence that tends to either positive or negative infinity is divergent! We avoid saying this kind of sequence "converges" to infinity to for precisely this reason.
Algebraic Limit Theorem
The Algebraic Limit Theorem is an important result that connects algebra on $\mathbb{R}$ to the limits of sequences in $\mathbb{R}$. Let $\{a_n\}$ and $\{b_n\}$ be sequences in $\mathbb{R}$ that converge to $p$ and $q$, respectively, and let $c \in \mathbb{R}.$ The algebraic limit theorem states the following:
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$\lim\limits_{n \rightarrow \infty} ca_n = c \lim\limits_{n \rightarrow \infty} a_n.$
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$\lim\limits_{n \rightarrow \infty} (a_n + b_n) = \lim\limits_{n \rightarrow \infty} a_n + \lim\limits_{n \rightarrow \infty} b_n.$
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$\lim\limits_{n \rightarrow \infty} (a_nb_n) = \left(\lim\limits_{n \rightarrow \infty} a_n \right) \left( \lim\limits_{n \rightarrow \infty} b_n \right).$
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$\lim\limits_{n \rightarrow \infty} \dfrac{1}{b_n} = \dfrac{1}{ \lim\limits_{n \rightarrow \infty} b_n }.$
- Corollary: $\lim\limits_{n \rightarrow \infty} \dfrac{a_n}{b_n} = \dfrac{\lim\limits_{n \rightarrow \infty}(a_n)}{\lim\limits_{n \rightarrow \infty}b_n}$.
Subsequential Limits
Let $\{a_{n_k}\}$ be a subsequence of $\{a_n\}$. If $\{a_{n_k}\}$ converges to a value $p$, then $p$ is a subsequential limit of $\{a_n\}$. Sequences may have one, many, or no subsequential limits. As is shown below, a sequence converges to $p$ if and only if all of its subsequences also converge to $p$.
Problems
Show that the sequence $\{a_n\}$ where $a_n = 1$ converges to $1$.
Pick some $\varepsilon > 0$. Then we'd like to find some $N \in \mathbb{N}$ such that $d(a_n, 1) < \varepsilon$ for all $n > N$. By definition, $a_n = 1$ for all $n \in \mathbb{N}$. Therefore $d(a_n, 1) = d(1, 1) = 0 < \varepsilon$. Thus we may take $N = 0$ and conclude $\{a_n\}$ converges to $1.$
Assume $\lim\limits_{n \rightarrow \infty}\{a_n\} = p$ in a discrete metric space $X$. Show that there is an $N \in \mathbb{N}$ such that $a_n = p$ for all $n > N.$
Since $\{a_n\}$ converges to $p$, then for every $\varepsilon > 0$ there is some $N \in \mathbb{N}$ such that $d(a_n, p) < 0$ for all $n > N$. Pick $\varepsilon = \frac{1}{2}.$ Since the discrete metric only takes on the values $0$ and $1$, it follows that $d(a_n, p) = 0$ for all $n > N$. By definition of a metric, it follows that $a_n = p$ for all $n > N$.
Show that the sequence $\left\{\frac{1}{n}\right\}$ converges to $0$ in $\mathbb{R}$.
Pick some positive $\varepsilon \in \mathbb{R}$. We must find an $N \in \mathbb{N}$ such that $d\left(\frac{1}{n}, 0\right) < \varepsilon$ for all $n > N.$ To do this, expand the metric $d$ to its algebraic definition and solve for $n$:
$ d\left(\dfrac{1}{n}, 0\right) < \varepsilon \\ \left|\dfrac{1}{n} - 0\right| < \varepsilon \\ \dfrac{1}{n} < \varepsilon \\ \dfrac{1}{\varepsilon} < n$
By the Archimedean property, there is an $N \in \mathbb{N}$ greater than $\dfrac{1}{\varepsilon}.$ The result follows.
Show that the sequence $\left\{\frac{1}{n}\right\}$ diverges in $\mathbb{R}^+$, i.e. the positive real numbers.
Proof by contradiction. Assume there is some value $\frac{1}{p} \in \mathbb{R}^+$ such that $\lim\limits_{n \rightarrow \infty}\left\{\frac{1}{n}\right\} = \frac{1}{p}$. Let $\varepsilon = \frac{1}{2p}$. Then there is some $N \in \mathbb{N}$ such that $d\left(\frac{1}{n}, \frac{1}{p}\right) < \frac{1}{2p}$ for all $n > N$. This means that $\left|\frac{1}{n} - \frac{1}{p} \right| < \frac{1}{2p}$. By the Archimedean principle, there is some $n > 4p$. But $\left|\frac{1}{n} - \frac{1}{p}\right| > \left|\frac{1}{4p} - \frac{1}{p}\right| = \frac{3}{4p} > \frac{1}{2p}$, which is a contradiction. Therefore $\left\{\frac{1}{n}\right\}$ does not converge in $\mathbb{R}^+$.
Limits are unique: Show that if $\{a_n\}$ converges to two points $p_1$ and $p_2$, then $p_1 = p_2$.
Assume $\{a_n\}$ converges to both $p_1$ and $p_2$ and that $p_1 \neq p_2$. Since $\{a_n\}$ converges to $p_1$, then for any $\varepsilon$ there is some $N$ such that $d(a_n, p_1) < \varepsilon$ for all $n > N$. Likewise, since $\{a_n\}$ also converges to $p_2$, then there is some $M$ such that $d(a_m, p_2) < \varepsilon$ for all $m > M$. Take $K = \text{max}(N, M)$. Then $d(a_k, p_1) < \varepsilon$ and $d(a_k, p_2) < \varepsilon$ for all $k > K$. Because $p_1 \neq p_2$, it follows that $q = d(p_1, p_2) > 0$. Select $\varepsilon = \frac{q}{4}$. By the triangle inequality, we see that
$ d(p_1, p_2) \leq d(p_1, a_k) + d(a_k, p_2) \\ d(p_1, p_2) < \varepsilon + \varepsilon \\ q < \dfrac{q}{4} + \dfrac{q}{4} \\ q < \dfrac{q}{2}$
However, this is a contradiction. Therefore $p_1 = p_2$ after all.
Show that if $\lim\limits_{n \rightarrow \infty} a_n = p$, then $\lim\limits_{n \rightarrow \infty} |a_n| = |p|.$
Assume $\lim\limits_{n \rightarrow \infty} a_n = p$. Then for any $\varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that $|a_n - p| < \varepsilon$ for all $n > \mathbb{N}$. Note that $\bigl||a_n| - |p|\bigr| \leq |a_n - p|$. Therefore $\bigl||a_n| - |p|\bigr| < \varepsilon$. Therefore $\lim\limits_{n \rightarrow \infty} |a_n| = |p|$.
Algebraic Limit Theorem 1 - Scalar Products: Show that $\lim\limits_{n \rightarrow \infty} ca_n = c \lim\limits_{n \rightarrow \infty} a_n$ where $\{a_n\}$ is a convergent sequence in $\mathbb{R}$ and $c \in \mathbb{R}.$
Let $c \in \mathbb{R}$ and let $\{a_n\}$ be a convergent sequence in $\mathbb{R}$ where $\lim\limits_{n \rightarrow \infty} a_n = p$. Then for every $\frac{\varepsilon}{|c|} \in \mathbb{R}^+$ there is some $N \in \mathbb{N}$ such that for all $n > N$, $d(a_n, p) < \frac{\varepsilon}{|c|}$. Then $|a_n - p| < \frac{\varepsilon}{|c|}$. Therefore $|c||a_n - p| < \varepsilon$, and so $|ca_n - cp| < \varepsilon$. Thus $d(ca_n, cp) < \varepsilon$, therefore $\lim\limits_{n \rightarrow \infty}ca_n = cp = c\lim\limits_{n \rightarrow \infty} a_n$.
Algebraic Limit Theorem 2 - Sums: Show that $\lim\limits_{n \rightarrow \infty} a_n + \lim\limits_{n \rightarrow \infty} = \lim\limits_{n \rightarrow \infty} (a_n + b_n)$ where $\{a_n\}$ and $\{b_n\}$ are convergent sequences in $\mathbb{R}$.
Let $\{a_n\}$ converge to $p$ and let $\{b_n\}$ converge to $q$. Then for any $\frac{\varepsilon}{2} \in \mathbb{R}^+$ there is an $N \in \mathbb{N}$ such that for all $n > N$, $d(a_n, p) < \frac{\varepsilon}{2}$, and there is an $M \in \mathbb{N}$ such that for all $m > M$, $d(b_m, q) < \frac{\varepsilon}{2}$. Take $K = \text{max}(N, M)$. Then for all $k > K$ it follows that $d(a_k, p) < \frac{\varepsilon}{2}$ and $d(b_k, q) < \frac{\varepsilon}{2}$. It follows that $|a_k - p| < \frac{\varepsilon}{2}$ and $|b_k - q| < \frac{\varepsilon}{2}$. Summing each inequality gives us $|(a_k - p) + (b_k - q)| < |a_k - p| + |b_k - q| < \varepsilon$.
Observe that $d(a_n + b_n, p + q) = |(a_n + b_n) - (p + q)|.$ Rearranging the righthand side gives $|(a_n - p) + (b_n - q)|$. By the properties of absolute value, we see that $|(a_n - p) + (b_n - q)| < |a_n - p| + |b_n - q| < \varepsilon$. Thus $d(a_n + b_n, p + q) < \varepsilon$. Therefore $\lim\limits_{n \rightarrow \infty} (a_n + b_n) = \lim\limits_{n \rightarrow \infty} a_n + \lim\limits_{n \rightarrow \infty} b_n$.
Algebraic Limit Theorem 3 - Products: Show that $\lim\limits_{n \rightarrow \infty} a_nb_n = \left(\lim\limits_{n \rightarrow \infty} a_n \right)\left(\lim\limits_{n \rightarrow \infty} b_n \right)$.
We want to connect the limit of the products to the product of the limits. Since the limit of the products is what we are trying to solve for, we must start by examining the product of the limits.
First, let $\{a_n\}$ converge to $p$ and let $\{b_n\}$ converge to $q$, and consider $(a_n - p)$ and $(b_n - q)$, the expressions we evaluate for each individual limit. Taking their product gives
$(a_n - p)(b_n - q) = a_nb_n - qa_n - pb_n + pq $
We can now solve for $a_nb_n - pq$, the expression we would like to evaluate for the product of the limits:
$a_nb_n - pq = (a_n - p)(b_n - q) + qa_n + pb_n - 2pq$
We rearrange the righthand side to be fully in terms of differences between sequences and their limits:
$ a_nb_n - pq = (a_n - p)(b_n - q) + qa_n + pb_n - 2pq \\ a_nb_n - pq = (a_n - p)(b_n - q) + qa_n - pq + pb_n - pq \\ a_nb_n - pq = (a_n - p)(b_n - q) + q(a_n - p) + p(b_n - q) $
We now take the absolute value of both sides and apply inequalities to isolate the limit terms:
$ |a_nb_n - pq| = |(a_n - p)(b_n - q) + q(a_n - p) + p(b_n - q)| \\ |a_nb_n - pq| \leq |(a_n - p)(b_n - q)| + |q(a_n - p)| + |p(b_n - q)| \\ |a_nb_n - pq| \leq |a_n - p||b_n - q| + |q||a_n - p| + |p||b_n - q| \\ $
Since $\{a_n\}$ converges to $p$ and $\{b_n\}$ converges to $q$, we can make the following claims:
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For every $\sqrt{\frac{\varepsilon}{2}} > 0$, there exists an $N \in \mathbb{N}$ such that $|a_n - p| < \sqrt{\frac{\varepsilon}{2}}.$ for all $n > N.$
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For every $\sqrt{\frac{\varepsilon}{2}} > 0$, there exists an $M \in \mathbb{N}$ such that $|b_m - q| < \sqrt{\frac{\varepsilon}{2}}$ for all $m > M.$
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For every $\dfrac{\varepsilon}{4|q|} > 0$, there exists a $U \in \mathbb{N}$ such that $|a_u - p| < \dfrac{\varepsilon}{4|q|}$ for all $u > U.$
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For every $\dfrac{\varepsilon}{4|p|} > 0$, there exists a $V \in \mathbb{N}$ such that $|b_v - q| < \dfrac{\varepsilon}{4|p|}$ for all $v > V.$
Take $K = \text{max}\{N, M, U, V\}$. Then we can substitute each value in for all $k > K$:
$ |a_kb_k - pq| < \left(\sqrt{\frac{\varepsilon}{2}}\right)\left(\sqrt{\frac{\varepsilon}{2}}\right) + |q|\left(\dfrac{\varepsilon}{4|q|}\right) + |p|\left(\dfrac{\varepsilon}{4|p|}\right) \\ |a_kb_k - pq| < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{4} + \dfrac{\varepsilon}{4} \\ |a_kb_k - pq| < \varepsilon $
Therefore $\lim\limits_{n \rightarrow \infty} a_nb_n = pq.$
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Algebraic Limit Theorem 4 - Inverses: Show that $\lim\limits_{n \rightarrow \infty} \dfrac{1}{a_n} = \dfrac{1}{\lim\limits_{n \rightarrow \infty}a_n}$ where $a_n \neq 0$ for all $n \in \mathbb{N}$ and $p \neq 0$.
Assume $\lim\limits_{n \rightarrow \infty} a_n = p$. We must find a way to connect the difference of the reciprocals to the difference of the original values. Consider the difference of the reciprocals:
$\dfrac{1}{a_n} - \dfrac{1}{p}$
Multiplying the first term by $\frac{p}{p}$ and the second term by $\frac{a_n}{a_n}$ brings out the term $p - a_n$ in the numerator:
$ \dfrac{1}{a_n} - \dfrac{1}{p} = \dfrac{p}{pa_n} - \dfrac{a_n}{pa_n} \\ \dfrac{1}{a_n} - \dfrac{1}{p} = \dfrac{p - a_n}{pa_n} \\ $
Taking the absolute value of both sides lets us use the $\varepsilon$ bound:
$ \left|\dfrac{1}{a_n} - \dfrac{1}{p}\right| = \left|\dfrac{p - a_n}{pa_n}\right| \\ \left|\dfrac{1}{a_n} - \dfrac{1}{p}\right| = \dfrac{|p - a_n|}{|pa_n|} \\ \left|\dfrac{1}{a_n} - \dfrac{1}{p}\right| = \dfrac{|a_n - p|}{|pa_n|} \\ \left|\dfrac{1}{a_n} - \dfrac{1}{p}\right| = \dfrac{|a_n - p|}{|p||a_n|}$
The righthand side contains the term $|a_n - p|$ in the numerator, which we know will converge to $0$. However, the denominator also contains an $a_n$ term in it. This is a problem, since solving for the limit of the reciprocal is the exact thing we are trying to do. However, if we can bound this $a_n$ term somehow, we will be able to replace it.
Since $\{a_n\}$ converges to $p$, for any $\varepsilon > 0$ there is some $N \in \mathbb{N}$ such that $|a_n - p| < \varepsilon$ for every $n > N$. Pick $\varepsilon = \frac{1}{2}|p|$. Then $|a_n - p| < \frac{1}{2}|p|$. However, rather than substitute this inequality in for the numerator, note that it implies $|a_n| > \frac{1}{2}|p|$. We substitute this value into the denominator instead:
$\left|\dfrac{1}{a_n} - \dfrac{1}{p}\right| < \dfrac{2|a_n - p|}{|p||p|} \\ \left|\dfrac{1}{a_n} - \dfrac{1}{p}\right| < \dfrac{2|a_n - p|}{|p|^2}$
Now pick $\frac{1}{2}|p|^2\varepsilon > 0$. Then there is an $M \in \mathbb{B}$ such that $|a_m - p| < \frac{1}{2}|p|^2\varepsilon$ for all $m > M.$ Take $K = \text{max}\{N, M\}$. For all $k > K$, the inequality then simplifies:
$\left|\dfrac{1}{a_n} - \dfrac{1}{p}\right| < \dfrac{2|p|^2\varepsilon}{2|p|^2} \\ \left|\dfrac{1}{a_n} - \dfrac{1}{p}\right| < \varepsilon$
Therefore $\lim\limits_{n \rightarrow \infty} \dfrac{1}{a_n} = \dfrac{1}{p}$.
Derive the corollary of the Algebraic Limit Theorem regarding products and inverses: If $\lim\limits_{n \rightarrow \infty} a_n = p$ and $\lim\limits_{n \rightarrow \infty} b_n = q$ and $b_n \neq 0$ for all $n \in \mathbb{N}$ and $q \neq 0$, then $\lim\limits_{n \rightarrow \infty} \frac{a_n}{b_n} = \frac{p}{q}.$
The result follows from the algebraic limit theorem:
$\lim\limits_{n \rightarrow \infty} \dfrac{a_n}{b_n} = \lim\limits_{n \rightarrow \infty} a_n \dfrac{1}{b_n} \\ \lim\limits_{n \rightarrow \infty} \dfrac{a_n}{b_n} = \left(\lim\limits_{n \rightarrow \infty} a_n\right)\left(\lim\limits_{n \rightarrow \infty} \dfrac{1}{b_n}\right) \\ \lim\limits_{n \rightarrow \infty} \dfrac{a_n}{b_n} = \dfrac{p}{q}$
Show that a sequence converges to $p$ if and only if all of its subsequences converge to $p$.
Consider the sequence $\{a_n\}$.
Assume $\{a_n\}$ converges to $p$, and let $\{a_{n_k}\}$ be any subsequence of $\{a_n\}$. Then for any $\varepsilon > 0$, there exists an $N$ such that $d(a_n, p) < \varepsilon$ for all $n > N$. It follows that $d(a_{n_k}, p) < \varepsilon$ for all $n_k > N$. Therefore $\{a_{n_k}\}$ converges to $p$.
Conversely, assume that all subsequences of $\{a_n\}$ converge to $p$. Because $\{a_n\}$ is a subsequence of itself, it therefore converges to $p$.
Give a counterexample to show that a sequence does not necessarily converge if all of its convergent subsequences converge to the same value.
Consider the sequence $\{0, 1, 0, 2, 0, 3, 0, 4, \ldots\}$. The only convergent subsequences are those that are restricted to the even numbered values (starting from $0$), all of which are $0$. These subsequences all converge to $0$, but the sequence diverges, as the odd-numbered values tend toward infinity.
Consider two sequences $\{a_n\}$ and $\{b_n\}$ that both converge to $p$. Construct the sequence $\{c_n\}$ by interleaving the values of $\{a_n\}$ and $\{b_n\}$ such that $\{c_n\} = \{a_0, b_0, a_1, b_1, a_2, b_2, \ldots\}$.
Does $\{c_n\}$ converge to $p$?
Yes.
Since $\{a_n\}$ and $\{b_n\}$ both converge to $p$, for any $\varepsilon > 0$, there are an $N$ and $M$ such that $d(a_n, p) < \varepsilon$ and $d(b_m, p) < \varepsilon$ for all $n > N$ and $m > M$. Take $K = 2\text{max}\{N, M\}$. Then $d(c_k, p) < \varepsilon$ for all $k > K$.