Convergent real sequences play a crucial role in mathematics for data science. In this post, we will first define real sequences and their convergence. We, then, list some basic properties of convergent real sequences. Finally, we give some examples to clarify the crucial points related to the convergence of real sequences.

### Convergent real sequences

#### Definition of real sequences

Suppose that \(A\) is a set. Informally speaking, a sequence in the set \(A\) is an ordered countable list of elements from the set \(A\). A finite sequence of length \(n\) has, for example, the following form: $$a_1, a_2, \dots, a_n,$$ where each \(a_i\) is an element of \(A\). An infinite sequence has, for example, the following form: $$a_1, a_2, \dots, a_n, \dots,$$ where, again, each \(a_i\) is an element of \(A\).

A sequence \((r_i)\) is real if \(r_i\) is a real number for each index \(i\).

**Example**. In the following, we give some examples of (infinite) real sequences:

- \(1, \sqrt{2}, \dots, \sqrt{n}, \dots\)
- \(\frac{1}{2}, \frac{2}{3}, \dots, \frac{n}{n+1}, \dots \)

In above, we defined (real) sequences informally. However, the formal definition of sequences is as follows:

**Definition**. A function \(f: D \rightarrow E\) is a sequence in the set \(E\) if \(D \subseteq \mathbb{Z}\).

By definition, a subsequence of a given sequence is a sequence that can be derived from the given sequence by deleting some or no elements without changing the order of the remaining elements. In other words, a subsequence of the sequence \(f: D \rightarrow E\), where \(D \subseteq \mathbb{Z}\), is any function of the form \(g: D’ \rightarrow E\), where \(D’ \subseteq D\) and \(g(x) = f(x)\) for each \(x \in D’\).

For example, \((a_{2n-1})\) is a subsequence of the sequence \((a_n)\).

#### Definition of convergent real sequences

Informally speaking, an infinite sequence $$a_1, a_2, \dots, a_n, \dots,$$ where each \(a_i\) is a real number is, by definition, convergent to a real number \(L\) whenever \(n\) becomes greater and greater, then \(a_n\) becomes closer and closer to \(L\).

For example, one may use a calculator, to informally prove that the sequence \(\frac{n}{n+1}\) is convergent to 1: $$a_1 = 0.5, a_2 = 0.666 \cdots, a_3 = 0.75$$ $$a_4 = 0.8, a_5 = 0.833 \cdots, a_6 = 0.857\cdots$$ $$a_7 = 0.875, a_8 = 0.888\cdots, a_9 = 0.9$$ and $$a_{1000000} = 0.999999000000.$$

The formal definition of convergent real sequences is as follows:

A (real) sequence $$a: \mathbb{N} \rightarrow \mathbb{R} \text{ defined by } n \mapsto a_n$$ is convergent to \(L\), denoted by \(\lim_{n \to +\infty} a_n = L\), if for each positive real number \(\epsilon\) there is a positive integer \(N\) such that \(n \geq N\) implies \(\vert a_n – L \vert < \epsilon\).

**Exercise**. Using the formal definition of convergent real sequences, prove that the constant sequence \(a_n = c\), where \(c\) is a real number is convergent to \(c\).

**Solution**. Let \(\epsilon\) be a positive real number. It is clear that for any positive integer \(N\), the assumption \(n \geq N\) implies that $$0 = \vert c – c \vert = \vert a_n – c \vert < \epsilon.$$

**Exercise**. Show that \(\lim_{n \to +\infty}\sqrt[n]{n} = 1.\)

#### Properties of convergent real sequences

In the following, we give the basic properties of convergent real sequences:

Assume that \((a_n)\) and \((b_n)\) are real sequences convergent to \(K\) and \(L\), respectively. Then, the following statements holds:

- The sequence \((a_n + b_n)\) is convergent to \(M+L\).
- The sequence \((a_n – b_n)\) is convergent to \(M-L\).
- The sequence \((a_n b_n)\) is convergent to \(M L\).
- If \(L \neq 0\), then the sequence \((a_n / b_n)\) is convergent to \(M / L\).
- If \(a_n > 0\) for each \(n \in \mathbb{N}\) and \(p\) is a positive real number, then \((a^p_n)\) is convergent to \(M^p\).

A real sequence \((a_n)\) is bounded above (below) if there is a fixed real number \(M\) such that for all positive integers \(n\), we have \(a_n < M\) (\(M < a_n\)).

A real sequence \((a_n)\) is bounded if there is a fixed positive real number \(M\) such that for all positive integers \(n\), we have \(\vert a_n \vert < M\).

**Exercise**. Prove that a real sequence \((a_n)\) is bounded if and only if \((a_n)\) is bounded above and below.

**Result**. Convergent real sequences are bounded.

Let \(\lim_{n \to +\infty} a_n = L\). Then, for \(\epsilon = 1\), there is a positive integer \(N\) such that for each \(n \geq N\), we have $$L – 1 < a_n < L+1.$$ On the other hand, each \(a_n\) for \(n \leq N\) is bounded above and below because $$\min \{a_n : n \leq N \} \leq a_i $$ and $$ a_i \leq \max \{a_n : n \leq N \},$$ for each \(i \leq N\). From all we said, we see that the sequence \((a_n)\) is bounded above and below. Hence, \((a_n)\) is bounded, as required.

**Exercise**. Let \(L\) be a real number. Prove that a real sequence \((a_n)\) is convergent to \(L\) if and only if each infinite subsequence of \((a_n)\) is convergent to \(L\).

**Exercise**. Let \(L\) be a real number. Prove that a real sequence \((a_n)\) is convergent to \(L\) if and only if the subsequences \((a_{2n-1})\) and \((a_{2n})\) are convergent to \(L\).

#### Monotone real sequences

A real sequence \((a_n)\) is increasing if \(a_{n+1} \geq a_n\), for each \(n\). Also, a real sequence \((a_n)\) is decreasing if \(a_{n+1} \leq a_n\), for each \(n\). On the other hand, a real sequence \((a_n)\) is strictly increasing if \(a_{n+1} > a_n\), for each \(n\), and strictly decreasing if \(a_{n+1} < a_n\), for each \(n\).

A real sequence is monotone if it is either increasing or decreasing.

One of the most important properties of monotone real sequences is the following:

**Result** (the monotone convergence theorem). A monotone real sequence is convergent if and only if it is bounded.

In the previous section, we just proved that any convergent sequence is bounded. The proof of this statement that a monotone and bounded real sequence is convergent is based on this fact that each bounded above (below) subset of the set of real numbers has a least upper bound (a greatest lower bound). In fact, it is easy to see that an increasing (decreasing) and bounded above (below) real sequence approaches to its least upper bound (greatest lower bound).

**Remark**. One may use the monotone convergence theorem to prove the Bolzano-Weierstrass theorem.

**Exercise**. Let \(a\) be a real number with \(\vert a \vert < 1\). Prove that $$\lim_{n \to +\infty} a^n = 0.$$

**Exercise** (Convergence of geometric series). Let \(a\) be a real number with \(\vert a \vert < 1\). Prove that $$\lim_{n \to +\infty} \sum_{i=0}^{n} a^i $$ $$ \frac{1}{1-a}.$$

#### Some examples of convergent real sequences

In every calculus book, one may find examples of convergent sequences. Also, we have given some examples of convergent sequences above. However, for the convenience of the reader, we solved some interesting exercises on the convergence of real sequences in the following:

**Exercise**. Prove that \(\lim_{n \to +\infty} (\sqrt[k]{n+1} – \sqrt[k]{n}) = 0\).

**Solution**. Observe that $$ 0 < \sqrt[k]{n+1} – \sqrt[k]{n} < $$ $$ n^k \left( (1+1/n)^k – 1 \right) < $$ $$ n^k \left( (1+1/n) – 1 \right) = 1/n^{1-k}.$$ Since \(1/n^{1-k} \rightarrow 0\) as \(n \rightarrow +\infty\), we obtain the desired result.

**Exercise**. Let \(s_1 = \sqrt{2}\) and \(s_{n+1} = \sqrt{2+s_n}\). Prove that \(s_n = 2\cos(\pi/{2^{n+1}})\). Use this to show that the following statements hold:

- \(s_n < s_{n+1}\), for all \(n \in \mathbb{N}\).
- \(s_n < 2\), for all \(n \in \mathbb{N}\).
- $$\sqrt{2+\sqrt{2+\sqrt{2+ \cdots}}} = $$ $$\lim_{n \to +\infty} s_n = 2.$$

The proof goes as follows:

It is obvious that $$\sqrt{2} = s_1 = 2 \cos(\pi/4).$$ Now, assume that $$s_n = 2\cos(\pi/{2^{n+1}}).$$ We need to prove that $$s_{n+1} = 2\cos(\pi/{2^{n+2}}).$$ From elementary trigonometry, we know that $$\cos(2x) = 2\cos^2 x – 1.$$ Now, observe that $$s_{n+1} = \sqrt{2+s_n} = $$ $$\sqrt{2+2\cos(\pi/{2^{n+1}})} = $$ $$ \sqrt{2+4\cos^2 (\pi/2^{n+2}) – 2} = $$ $$2\cos(\pi/{2^{n+2}}).$$

Now, it is clear that \(s_n < s_{n+1} < 2\), for each \(n \in \mathbb{N}\), and $$\lim_{n \to +\infty} s_n = $$ $$\lim_{n \to +\infty} 2\cos(\pi/{2^{n+1}}) = $$ $$ 2\cos(0) = 2.$$

**Exercise**. Let \(x_1 = a\), \(x_2 = b\), and \(x_{n+2} = \displaystyle \frac{x_{n+1} + x_n}{2}\), where \(a\) and \(b\) are real numbers and \(n\) is an arbitrary positive integer. Evaluate \(\lim_{n \to +\infty} x_n.\)

**Solution**. Observe that $$x_{n+2} – x_{n+1} = -\frac{1}{2} \cdot (x_{n+1} – x_n).$$ If we set \(s_n = x_{n+1} – x_n\), we have $$s_n = (-1/2)^{n-1} \cdot (b-a).$$ Also, by computing the limit of the following geometric series (see above exercise on the convergence of geometric series), we have $$\lim_{n \to +\infty} (x_{n+1} – x_1) = \sum_{n=1}^{+\infty} s_n$$ $$ = \frac{b-a}{1 – (-1/2)} = (2/3)(b-a).$$ Finally, it is evident that $$\lim_{n \to +\infty} x_{n} = \frac{a+2b}{3}.$$

#### The Fibonacci-Pingala sequence

**Exercise**. By definition, the Fibonacci-Pingala sequence is defined as follows:

$$F_1 = 1 \text{ and } F_2 = 1$$ and $$F_{n+2} = F_{n+1} + F_n \text{ for each } n \in \mathbb{N}.$$ Prove that $$\lim_{n \to +\infty} \frac{F_{n+1}}{F_n} = \frac{1+\sqrt{5}}{2}.$$

For different types of numbers see the post on types of numbers.

**Remark**. A sequence is divergent if it is not convergent.