An infinite sequence is a sequence $a_1, a_2, a_3, \dots, a_n, \dots$ having an infinite number of terms. The indicated sum of its terms is called an infinite series associated with the sequence:
$$a_1 + a_2 + a_3 + \dots + a_n + \dots = \sum_{k=1}^{\infty} a_k$$In this chapter, we study special types of infinite series, including the Binomial Series, Infinite Geometric Series, Exponential Series, and Logarithmic Series.
When the index $m$ is a negative integer or a fraction, the expansion of $(1+x)^m$ yields an infinite series called the Binomial Series.
The formula:
$$(1+x)^m = 1 + mx + \frac{m(m-1)}{1 \cdot 2}x^2 + \frac{m(m-1)(m-2)}{1 \cdot 2 \cdot 3}x^3 + \dots$$holds if and only if $|x| < 1$.
To expand $(a+b)^m$ when $m$ is negative or fractional, we factor out $a^m$:
$$(a+b)^m = a^m \left(1 + \frac{b}{a}\right)^m = a^m \left[ 1 + m\left(\frac{b}{a}\right) + \frac{m(m-1)}{1 \cdot 2}\left(\frac{b}{a}\right)^2 + \dots \right]$$This expansion is valid if and only if $\left|\frac{b}{a}\right| < 1$, which is equivalent to $|b| < |a|$.
The general term ($(r+1)$-th term) in the expansion of $(a+b)^m$ is given by:
$$T_{r+1} = \frac{m(m-1)(m-2)\dots(m-r+1)}{1 \cdot 2 \cdot 3 \dots r} a^{m-r} b^r$$Assuming $|x| < 1$, the following standard expansions are highly useful:
| Binomial Expression | Infinite Series Expansion |
|---|---|
| $(1+x)^{-1}$ | $1 - x + x^2 - x^3 + x^4 - \dots$ |
| $(1-x)^{-1}$ | $1 + x + x^2 + x^3 + x^4 + \dots$ |
| $(1+x)^{-2}$ | $1 - 2x + 3x^2 - 4x^3 + \dots$ |
| $(1-x)^{-2}$ | $1 + 2x + 3x^2 + 4x^3 + \dots$ |
Mathematical Problem Statement: Expand $\left(1 - \frac{x}{2}\right)^{-\frac{1}{2}}$ when $|x| < 2$.
Step-by-Step Algorithmic Solution Trace:
- Identify the parameters: let $y = -\frac{x}{2}$ and $m = -\frac{1}{2}$. Since $|x| < 2$, we have $|y| = \left|-\frac{x}{2}\right| < 1$, so the binomial expansion is valid.
- Apply the binomial series formula:$$\left(1 - \frac{x}{2}\right)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)\left(-\frac{x}{2}\right) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{1 \cdot 2}\left(-\frac{x}{2}\right)^2 + \dots$$
- Simplify each term:
- First term: $1$
- Second term: $\frac{x}{4}$
- Third term: $\frac{3/4}{2} \cdot \frac{x^2}{4} = \frac{3x^2}{32}$
Final Evaluation Output: $\left(1 - \frac{x}{2}\right)^{-\frac{1}{2}} = 1 + \frac{x}{4} + \frac{3x^2}{32} + \dots$
An infinite geometric series is of the form $a + ar + ar^2 + \dots + ar^{n-1} + \dots$, where $a$ is the first term and $r$ is the common ratio.
The sum of the first $n$ terms of a geometric progression (G.P.) is given by:
$$S_n = \frac{a(1-r^n)}{1-r} = \frac{a}{1-r} - \frac{ar^n}{1-r}$$If the numerical value of the common ratio $r$ is less than 1 (i.e., $|r| < 1$):
Mathematical Problem Statement: Find the sum to infinity of the G.P.: $-\frac{5}{4}, \frac{5}{16}, -\frac{5}{64}, \dots$
Step-by-Step Algorithmic Solution Trace:
- Identify the first term: $a = -\frac{5}{4}$.
- Identify the common ratio: $r = \frac{5/16}{-5/4} = -\frac{1}{4}$.
- Verify convergence: $|r| = \left|-\frac{1}{4}\right| = \frac{1}{4} < 1$, so the sum to infinity exists.
- Apply the infinite sum formula:$$S = \frac{a}{1-r} = \frac{-5/4}{1 - (-1/4)} = \frac{-5/4}{5/4} = -1$$
Final Evaluation Output: $S = -1$
The exponential series is an infinite series of numbers whose sum defines the mathematical constant $e$.
Historical Chronicle: Leonhard Euler (1707 – 1783)
The constant $e$ is defined by the infinite series:
$$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots$$We can estimate the value of $e$ by comparing its terms with a geometric series:
For any real variable $x$, the exponential function $e^x$ can be expanded as:
$$e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!} + \dots$$Mathematical Problem Statement: Find the coefficient of $x^2$ in the expansion of $e^{2x+3}$ as a series in powers of $x$.
Step-by-Step Algorithmic Solution Trace:
- Express $e^{2x+3}$ as a product: $e^{2x+3} = e^3 \cdot e^{2x}$.
- Expand $e^{2x}$ using the general exponential series:$$e^{2x} = 1 + \frac{2x}{1!} + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots$$
- Multiply by $e^3$:$$e^{2x+3} = e^3 \left[ 1 + \frac{2x}{1!} + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \dots \right]$$
- Extract the coefficient of $x^2$:$$\text{Coefficient of } x^2 = e^3 \cdot \frac{4}{2!} = 2e^3$$
Final Evaluation Output: $2e^3$
Mathematical Problem Statement: Find the value of $e^2$, rounded off to one decimal place.
Step-by-Step Algorithmic Solution Trace:
- Substitute $x = 2$ into the exponential series:$$e^2 = 1 + \frac{2}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \frac{2^5}{5!} + \frac{2^6}{6!} + \dots$$
- Calculate the sum of the first seven terms:$$e^2 \ge 1 + 2 + 2 + \frac{4}{3} + \frac{2}{3} + \frac{4}{15} + \frac{4}{45} \approx 7.355$$
- Establish an upper bound using a geometric progression for the tail terms:$$e^2 < 7 + \frac{4}{15}\left(1 + \frac{1}{3} + \left(\frac{1}{3}\right)^2 + \dots\right) = 7 + \frac{4}{15}\left(\frac{1}{1 - 1/3}\right) = 7 + \frac{2}{5} = 7.4$$
- Since $7.355 \le e^2 < 7.4$, rounding to one decimal place yields 7.4.
Final Evaluation Output: $7.4$
The logarithmic series is an infinite series expansion for natural logarithms.
If $|x| < 1$, then:
$$\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$$Special Case ($x = 1$): The expansion is also valid for $x = 1$, yielding:
$$\log_e 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$$Mathematical Problem Statement: If $\alpha, \beta$ are the roots of the quadratic equation $x^2 - px + q = 0$, prove that:$$\log_e(1 + px + qx^2) = (\alpha + \beta)x - \frac{\alpha^2 + \beta^2}{2}x^2 + \frac{\alpha^3 + \beta^3}{3}x^3 - \dots$$
Step-by-Step Algorithmic Solution Trace:
- Recall the relations between roots and coefficients: $\alpha + \beta = p$ and $\alpha\beta = q$.
- Express the quadratic term inside the logarithm in factored form:$$1 + px + qx^2 = 1 + (\alpha + \beta)x + \alpha\beta x^2 = (1 + \alpha x)(1 + \beta x)$$
- Apply logarithmic properties:$$\log_e(1 + px + qx^2) = \log_e[(1 + \alpha x)(1 + \beta x)] = \log_e(1 + \alpha x) + \log_e(1 + \beta x)$$
- Expand both terms using the logarithmic series (assuming $|\alpha x| < 1$ and $|\beta x| < 1$):$$\log_e(1 + \alpha x) = \alpha x - \frac{\alpha^2 x^2}{2} + \frac{\alpha^3 x^3}{3} - \dots$$$$\log_e(1 + \beta x) = \beta x - \frac{\beta^2 x^2}{2} + \frac{\beta^3 x^3}{3} - \dots$$
- Group the terms by powers of $x$:$$\log_e(1 + px + qx^2) = (\alpha + \beta)x - \frac{\alpha^2 + \beta^2}{2}x^2 + \frac{\alpha^3 + \beta^3}{3}x^3 - \dots$$
Final Evaluation Output: The identity is proved.