18.1 Arithmetic & Geometric Sequences, Finite Sums & Infinite Series Convergence

18.1 Arithmetic & Geometric Sequences, Finite Sums & Infinite Series Convergence

Sequence Foundations: Explicit vs. Recursive Formulations

A sequence is formally defined as a function whose domain is the set of positive integers $\mathbb{Z}^+ = \{1, 2, 3, \dots\}$ (or non-negative integers $\mathbb{N}_0 = \{0, 1, 2, \dots\}$), with range values corresponding to the terms of the sequence $a_1, a_2, a_3, \dots, a_n$. Sequences are characterized algebraically through two primary modalities:

1. Explicit Formulas: An explicit formula expresses the $n$-th term directly as a function of its index: $a_n = f(n)$. This representation allows instantaneous determination of any arbitrary term without prior calculation of preceding terms. For example, if $a_n = 3n^2 - 5$, the 15th term is directly $a_{15} = 3(15)^2 - 5 = 670$.
2. Recursive Formulas: A recursive specification defines the $n$-th term in relation to one or more preceding terms: $a_n = g(a_{n-1})$, requiring one or more initial boundary conditions (such as $a_1$). For instance, the system $a_1 = 4, a_n = 3a_{n-1} - 2$ generates successive terms iteratively: $a_2 = 3(4) - 2 = 10$, $a_3 = 3(10) - 2 = 28$.

On the FTCE Mathematics 6-12 examination, translating between recursive relations and closed-form explicit expressions is a core analytical competency.

---

Arithmetic Sequences & Finite Arithmetic Series

An arithmetic sequence is defined by a constant difference $d$ between any term and its predecessor, known as the common difference:

$$d = a_n - a_{n-1} \quad \text{for all } n \ge 2$$

The recursive definition is $a_n = a_{n-1} + d$ with given $a_1$. Unfolding this relation across successive indices yields the general explicit formula:

$$a_n = a_1 + (n - 1)d$$

Given two arbitrary terms $a_j$ and $a_k$ (with $k > j$), the common difference is uniquely calculated by $d = \frac{a_k - a_j}{k - j}$.

The sum of the first $n$ terms of an arithmetic sequence constitutes a finite arithmetic series, denoted $S_n = \sum_{k=1}^n a_k$. Deriving this sum by writing the terms forward and backward and adding the paired equations produces $2S_n = n(a_1 + a_n)$, leading directly to the classic summation formulas:

$$S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}[2a_1 + (n - 1)d]$$

The quantity $\frac{a_1 + a_n}{2}$ represents the arithmetic mean of the first and last terms; thus, the sum is simply the number of terms multiplied by the average term value.

---

Geometric Sequences & Finite Geometric Series

A geometric sequence is defined by a constant non-zero multiplier $r$ between consecutive terms, designated the common ratio:

$$r = \frac{a_n}{a_{n-1}} \quad \text{for all } n \ge 2 \quad (a_{n-1} \neq 0)$$

The recursive definition is $a_n = r \cdot a_{n-1}$ with initial term $a_1 \neq 0$. Repeated multiplication gives the explicit closed form:

$$a_n = a_1 r^{n-1}$$

For two terms $a_j$ and $a_k$, the terms satisfy $r^{k-j} = \frac{a_k}{a_j}$.

A finite geometric series is the sum of the first $n$ terms: $S_n = \sum_{k=1}^n a_1 r^{k-1} = a_1 + a_1 r + a_1 r^2 + \dots + a_1 r^{n-1}$. Multiplying $S_n$ by $r$ and subtracting the product from $S_n$ results in intermediate algebraic cancellation:

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

Dividing by $(1 - r)$ establishes the finite sum formula:

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

If $r = 1$, each term equals $a_1$, and the sum collapses to $S_n = n a_1$.

---

Infinite Geometric Series & Convergence Criteria

An infinite geometric series takes the form $\sum_{k=1}^\infty a_1 r^{k-1} = a_1 + a_1 r + a_1 r^2 + \dots$. The convergence of this infinite series is governed by the limiting behavior of the sequence of partial sums $\{S_n\}$ as $n \to \infty$:

$$S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{a_1(1 - r^n)}{1 - r}$$

Because $\lim_{n \to \infty} r^n = 0$ if and only if $|r| < 1$, the infinite series converges to a finite sum under the strict condition $|r| < 1$:

$$S = \frac{a_1}{1 - r} \quad \text{for } |r| < 1$$

If $|r| \ge 1$ (with $a_1 \neq 0$), $\lim_{n \to \infty} a_n \neq 0$, violating the Divergence Test ($n$-th Term Test for Divergence). Consequently, the series diverges, meaning its sum does not exist as a finite real number.

---

Repeating Decimals as Infinite Geometric Series

A direct pedagogical application of infinite geometric series is converting repeating decimals into simplified rational fractions $\frac{p}{q}$ ($p, q \in \mathbb{Z}, q \neq 0$). Every repeating decimal consists of an initial non-repeating part plus an infinite series whose common ratio is $r = 10^{-m}$, where $m$ is the period length of the repeating block. For example, consider $0.\overline{45} = 0.454545\dots$:

$$0.\overline{45} = \frac{45}{100} + \frac{45}{10,000} + \frac{45}{1,000,000} + \dots$$

Here, the first term is $a_1 = \frac{45}{100}$ and the common ratio is $r = \frac{1}{100} = 0.01$. Because $|r| = 0.01 < 1$, the infinite sum evaluates to:

$$S = \frac{\frac{45}{100}}{1 - \frac{1}{100}} = \frac{\frac{45}{100}}{\frac{99}{100}} = \frac{45}{99} = \frac{5}{11}$$

---

Sigma Notation Manipulation & Linearity Properties

Sigma notation provides a concise algebraic framework for summations: $\sum_{k=m}^n a_k = a_m + a_{m+1} + \dots + a_n$. The total number of terms evaluated is $(n - m + 1)$. Essential properties include:

1. Linearity of Summation: $\sum_{k=m}^n (c a_k + b_k) = c \sum_{k=m}^n a_k + \sum_{k=m}^n b_k$, where $c$ is any constant.
2. Sum of a Constant: $\sum_{k=1}^n c = c \cdot n$.
3. Index Shifting: Shifting the summation counter by an integer $p$ alters the limits while maintaining equivalent terms: $\sum_{k=m}^n a_k = \sum_{j=m+p}^{n+p} a_{j-p}$.
4. Range Splitting: $\sum_{k=1}^n a_k = \sum_{k=1}^m a_k + \sum_{k=m+1}^n a_k$ for $1 \le m < n$.

---

Sequence and Series Formulas Summary

---

Worked Exemplar: Parameterized Infinite Geometric Series

Problem: Determine all values of $x$ for which the infinite geometric series $\sum_{k=1}^\infty 6 \left(\frac{3x - 4}{5}\right)^k$ converges. Express its sum $S(x)$ as a simplified rational function, and evaluate $S(1)$.

Step 1: Identify Initial Term and Common Ratio. Expand the first few terms of the series:

$$\sum_{k=1}^\infty 6 \left(\frac{3x - 4}{5}\right)^k = 6\left(\frac{3x - 4}{5}\right) + 6\left(\frac{3x - 4}{5}\right)^2 + 6\left(\frac{3x - 4}{5}\right)^3 + \dots$$

The first term is $a_1 = 6\left(\frac{3x - 4}{5}\right) = \frac{18x - 24}{5}$. The common ratio between consecutive terms is $r = \frac{3x - 4}{5}$.

Step 2: Determine Interval of Convergence. An infinite geometric series converges if and only if $|r| < 1$:

$$\left|\frac{3x - 4}{5}\right| < 1 \iff -1 < \frac{3x - 4}{5} < 1$$

Multiply through by 5:

$$-5 < 3x - 4 < 5$$

Add 4 to all expressions:

$$-1 < 3x < 9$$

Divide by 3:

$$-\frac{1}{3} < x < 3$$

Thus, the series converges strictly on the open interval $x \in \left(-\frac{1}{3}, 3\right)$.

Step 3: Construct Rational Sum Function $S(x)$. Apply the infinite sum formula $S = \frac{a_1}{1 - r}$:

$$S(x) = \frac{\frac{6(3x - 4)}{5}}{1 - \frac{3x - 4}{5}} = \frac{\frac{6(3x - 4)}{5}}{\frac{5 - (3x - 4)}{5}} = \frac{6(3x - 4)}{5 - 3x + 4} = \frac{18x - 24}{9 - 3x} = \frac{6(3x - 4)}{3(3 - x)} = \frac{2(3x - 4)}{3 - x}$$

Step 4: Evaluate the Sum at $x = 1$. Because $x = 1 \in \left(-\frac{1}{3}, 3\right)$, substitute directly into $S(x)$:

$$S(1) = \frac{2(3(1) - 4)}{3 - 1} = \frac{2(-1)}{2} = -1$$

Verifying with original series parameters at $x = 1$:

$$r = \frac{3(1) - 4}{5} = -\frac{1}{5}, \quad a_1 = 6\left(-\frac{1}{5}\right) = -\frac{6}{5}$$ $$S = \frac{-\frac{6}{5}}{1 - \left(-\frac{1}{5}\right)} = \frac{-\frac{6}{5}}{\frac{6}{5}} = -1$$

The derived closed-form function confirms the numerical sum.