Series and Sequences Part 1

What Is a Sequence?

In mathematics, sequences have found a place in modern mathematical analysis as systems in which functions are ordered according to a particular rule. Sequences are viewed as structures in which functions are arranged in a definite order. Many examples appeared in antiquity, such as the sequence found in the Rhind Papyrus among questions concerning division in arithmetic. The ancient Babylonians regarded sequences as geometric progressions used in agriculture and astronomy. The Greeks treated sequences and their sums—that is, series—as philosophical subjects. Zeno's paradoxes attempted to explain how a sequence formed by repeatedly halving a distance could yield a finite result even though the sequence contained infinitely many terms. Archimedes summed a geometric sequence while finding the area of a parabolic segment. As we can see, many civilizations throughout history used sequences both in agriculture and in exploring the heavens. In this article, however, we will approach the subject neither agriculturally nor astronomically, but entirely mathematically. Through numerous examples and proofs, you will come to understand what sequences are and what they are not.

Definition of Sequences

Definition
Any function whose domain is the positive integers ($\mathbb{Z}^+$) is called a sequence.

$$f(1) = a_1, \quad f(2) = a_2, \quad \dots, \quad f(n) = a_n$$

Here, the number $a_n$ corresponding to the natural number $n$ is called the general term of the sequence.

Example
$a_n = \frac{1}{n} + \frac{1}{n^3}$

denotes the sequence whose terms include

$a_1 = 2, \quad a_2 = \frac{5}{8}$

Example
Find the fourth term of the sequence
$\left( \frac{n^6}{(n+1)^2} \right)$

Solution
$a_4 = \frac{4^6}{(4+1)^2} = \frac{2^4}{25} = a_4$

is obtained.

Types of Sequences and Their General Forms

Definition
If there exists a $d \in \mathbb{R}$ such that
$a_{n+1} - a_n = d$

then $(a_n)$ is called an arithmetic sequence, and $d$ is called the common difference.

Example
Show that
$a_n = 6n + 11$

is an arithmetic sequence.

Solution
$a_n = 6n + 11$

$a_{n+1} = 6(n+1) + 11$

and

$a_{n+1} - a_n = 6 \in \mathbb{R}$

so $(a_n)$ is an arithmetic sequence.

Now let us find the sum of the first $n$ terms of the arithmetic sequence

$$a_n = kn + t$$

Let

$$S_n = a_1 + a_2 + a_3 + \cdots + a_n$$ $$a_1 = k + t$$ $$a_2 = 2k + t$$ $$a_3 = 3k + t$$ $$\vdots$$ $$a_n = kn + t$$

It follows that

$$S_n = \frac{n}{2}\left[k(n+1) + 2t\right]$$

We also obtain

$$S_n = \frac{n}{2}(a_1 + a_n)$$

Definition
If there exists an $r \in \mathbb{R}$ such that
$\frac{a_{n+1}}{a_n} = r$

then $(a_n)$ is a geometric sequence.

Example
Show that
$a_n = 3 \cdot 2^n$

is a geometric sequence.

Solution
$a_n = 3 \cdot 2^n$

$a_{n+1} = 3 \cdot 2^{n+1} = 6 \cdot 2^n$

Therefore,

$\frac{a_{n+1}}{a_n} = \frac{3 \cdot 2^{n+1}}{3 \cdot 2^n} = 2$

so $(a_n)$ is a geometric sequence, with common ratio

$r = 2$

Now let us find the sum of the first $n$ terms of a geometric sequence.

If

$$a_{n+1} = r a_n$$

then

$$a_2 = r a_1$$ $$a_3 = r a_2 = r^2 a_1$$ $$a_4 = r a_3 = r^3 a_1$$ $$\vdots$$ $$a_n = r^{n-1} a_1$$

Thus,

$$S_n = a_1 + a_2 + a_3 + \cdots + a_n$$ $$S_n = a_1 + a_1 r + a_1 r^2 + \cdots + a_1 r^{n-1}$$

Hence,

$$S_n = a_1\left(1 + r + r^2 + \cdots + r^{n-1}\right)$$

Let

$$S_n = a_1\left(1 + r + r^2 + \cdots + r^{n-1}\right)$$

Then

$$r S_n = a_1 r + a_1 r^2 + \cdots + a_1 r^n$$

Subtracting one side from the other gives

$$(1 - r)S_n = a_1 - a_1 r^n$$

Therefore,

$$S_n = \frac{a_1(1 - r^n)}{1 - r}$$

Monotonicity

Definition
$a_n < a_{n+1} \iff (a_n) \ \text{artan}$

$a_n > a_{n+1} \iff (a_n) \ \text{azalan}$

$a_n \leq a_{n+1} \iff (a_n) \ \text{azalmayan}$

$a_n \geq a_{n+1} \iff (a_n) \ \text{artmayan}$

Increasing or decreasing sequences are called monotone sequences.

Example
Examine the monotonicity of the sequence
$\left( \frac{2^n}{(n+1)!} \right)$

Solution
$\frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{(n+2)!} \cdot \frac{(n+1)!}{2^n} = \frac{2}{n+2}$

Since

$n + 2 > 2$

we have

$\frac{2}{n+2} < 1$

and therefore

$\frac{a_{n+1}}{a_n} < 1$

It follows that

$a_{n+1} < a_n$

Thus $(a_n)$ is decreasing.

Definition of Boundedness for Sequences

Definition
If there exists an $M \in \mathbb{R}$ such that, for every $n \in \mathbb{N}$,

$a_n \leq M$

then the sequence $a_n$ is said to be bounded above.
Similarly, if there exists an $m \in \mathbb{R}$ such that

$a_n \geq m$

then the sequence is bounded below.

Example
Examine whether the sequence
$\left( (-1)^n \frac{n}{n+1} \right)$

is bounded.

Solution
$-1 < -\frac{n}{n+1} \leq (-1)^n \frac{n}{n+1} \leq \frac{n}{n+1} < 1$

Hence the sequence is bounded both below and above; that is, it is bounded.

Convergence of Sequences

If, apart from finitely many terms, all terms of a sequence $(a_n)$ lie in every neighborhood of a number $a$, then the sequence is said to converge to $a$, or $a$ is called the limit of $(a_n)$.

$$\lim a_n = a$$

or

$$(a_n) \to a$$

A sequence is either convergent or divergent.

If, for every $\varepsilon > 0$, there exists a suitable natural number $N$ such that for all $n \geq N$,

$$|a_n - L| < \varepsilon$$

then

$$\lim a_n = L$$

Example
Show that the sequence
$a_n = \frac{1}{n}$

converges to 0.

Solution
Let
$L = 0$

The inequality

$|a_n - L| < \varepsilon$

becomes

$\left| \frac{1}{n} - 0 \right| < \varepsilon$

Thus,

$\frac{1}{n} < \varepsilon$

is required. This means

$n > \frac{1}{\varepsilon}$

Therefore, if

$N = \left[ \frac{1}{\varepsilon} \right]$

is chosen, then for

$n \geq N$

we have

$\left| \frac{1}{n} \right| < \varepsilon$

Consequently,

$(a_n) \to 0$

Limits of Sequences and Their Properties

Let

$$\lim a_n = a, \quad \lim b_n = b, \quad \delta \in \mathbb{R}$$

Then:

1. $\lim(a_n + b_n) = a + b$

2. $\lim(a_n b_n) = ab$

3. If $b_n \neq 0, \quad b \neq 0$, then

$$\lim \frac{a_n}{b_n} = \frac{a}{b}$$

4. $\lim(\delta a_n) = \delta a$

5. If $\lim a_n = a$, then

$$\lim \frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} = a$$

Proof

Suppose

$$\lim(a_n + b_n) = a + b$$

Set

$$a_n + b_n = c_n$$

If

$$\lim c_n = a + b$$

then for every $\varepsilon > 0$, there exists an $N > 0$ such that whenever

$$n > N$$

we have

$$|c_n - (a + b)| < \varepsilon$$

That is,

$$|a_n + b_n - a - b| < \varepsilon$$

From this,

$$|(a_n - a) + (b_n - b)| \leq |a_n - a| + |b_n - b|$$

Since $a_n \to a$, for every $\varepsilon_1 > 0$, there exists an $N_1$ such that when

$$n > N_1$$

we have

$$|a_n - a| < \varepsilon_1$$

Similarly, because $b_n \to b$, for every $\varepsilon_2 > 0$, there exists an $N_2$ such that when

$$n > N_2$$

we have

$$|b_n - b| < \varepsilon_2$$

Choosing

$$\varepsilon_1 = \frac{\varepsilon}{2}, \quad \varepsilon_2 = \frac{\varepsilon}{2}$$

and taking

$$N = \max(N_1, N_2)$$

completes the proof.

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Suppose

$$(a_n) \to a$$

We will show that

$$\lim \frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} = a$$

Proof

$$|a_n - a| < \varepsilon$$

can be ensured by choosing

$$n > N_0$$

and a suitable $\varepsilon > 0$.

Split the sum into two parts:

$$\lim \frac{a_1 + a_2 + \cdots + a_{N_0}}{n} + \lim \frac{a_{N_0+1} + a_{N_0+2} + \cdots + a_n}{n}$$

The first part contains finitely many terms, so

$$\lim \frac{a_1 + a_2 + \cdots + a_{N_0}}{n} = 0$$

Therefore,

$$\lim \frac{a_{N_0+1} + a_{N_0+2} + \cdots + a_n}{n} = a$$

This portion contains approximately $n - N_0$ terms.
Treating each term as approximately $a$, the sum is approximately

$$a(n - N_0)$$

Thus,

$$\frac{a(n - N_0)}{n} = a\left(1 - \frac{N_0}{n}\right)$$

Since

$$\lim_{n \to \infty} a\left(1 - \frac{N_0}{n}\right) = a(1 - 0) = a$$

we obtain

$$\boxed{\lim_{n \to \infty} \frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} = a}$$

The proofs of Theorems 2, 3, and 4 are left to the reader as exercises.

Conclusion

This article presented the history and properties of sequences, the general forms of geometric and arithmetic sequences, convergence, and increasing and decreasing behavior in an entirely formal manner, with formal proofs. I thought we would fit in much more, but let that be enough for this time. In the next article, I will discuss the convergence of geometric and arithmetic sequences at a much more abstract level. Take care.

References

• Balcı, M. (2020). Genel Matematik 1. Palme Yayınları.
• Apostol, T. M. Calculus, Vol. 1, 2nd ed., Wiley, 1967.