Problem 40 - Calculus: Early Transcendentals

Question

(a) Starting with the geometric series \( \sum_{n = 0}^{\infty} x^n, \) find the sum of the series \( \sum_{n = 1}^{\infty} nx^{n - 1} \mid x \mid < 1 \) (b) Find the sum of each of the following series. (i) \( \sum_{n = 1}^{\infty} nx^n, \mid x \mid < 1 \) (ii) \( \sum_{n = 1}^{\infty} \frac {n}{2^n} \) (c) Find the sum of each of the following series. (i) \( \sum_{n = 2}^{\infty} n(n - 1) x^n, \mid x \mid < 1 \) (ii) \( \sum_{n = 2}^{\infty} \frac {n^2 - n}{2^n} \) (iii) \( \sum_{n = 1}^{\infty} \frac {n^2}{2^n} \)

Step-by-Step Solution

Answer

a: \( \frac{1}{(1-x)^2} \); b(i): \( \frac{x}{(1-x)^2} \); b(ii): 2; c(i): \( \frac{x(x+1)}{(1-x)^3} \); c(ii): 6; c(iii): 8.

Step 1: Establish Geometric Series Formula

The geometric series formula is given by \( \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \) for \( |x| < 1 \). We use this as a starting point to derive other related series.

Step 2: Derive New Series for \( \sum_{n=1}^{\infty} nx^{n-1} \)

To derive the expression for \( \sum_{n=1}^{\infty} nx^{n-1} \), we differentiate the geometric series \( \sum_{n=0}^{\infty} x^n \) with respect to \( x \). This gives \( \sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2} \).

Step 3: Solve for \( \sum_{n=1}^{\infty} nx^n \)

From the series \( \sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2} \), multiply through by \( x \) to get \( \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2} \).

Step 4: Plug in for \( \sum_{n=1}^{\infty} \frac{n}{2^n} \)

Using \( x = \frac{1}{2} \) in \( \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2} \), we find \( \sum_{n=1}^{\infty} \frac{n}{2^n} = \frac{\frac{1}{2}}{(1-\frac{1}{2})^2} = 2 \).

Step 5: Analyze and Solve for \( \sum_{n=2}^{\infty} n(n-1)x^n \)

We differentiate \( \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2} \) with respect to \( x \) to find \( \sum_{n=2}^{\infty} n(n-1)x^n = \frac{x(x+1)}{(1-x)^3} \).

Step 6: Evaluate \( \sum_{n=2}^{\infty} \frac{n^2-n}{2^n} \)

We apply the formula from Step 5 with \( x = \frac{1}{2} \): \( \sum_{n=2}^{\infty} \frac{n^2-n}{2^n} = \frac{\frac{1}{2}(\frac{1}{2}+1)}{(1-\frac{1}{2})^3} = 6 \).

Step 7: Simplify and Find \( \sum_{n=1}^{\infty} \frac{n^2}{2^n} \)

Using the binomial expansion and adding constants, we express the problem in terms of steps 4 and 6: \( \sum_{n=1}^{\infty} \frac{n^2}{2^n} = 6 + 2 = 8 \).

Key Concepts

Infinite Series

Exploring an infinite series can unravel complex mathematical patterns in a simple way. An infinite series is essentially a sum of infinitely many terms that usually follow a specific rule or pattern. When dealing with infinite series, often the goal is to understand the behavior of the overall sum as the number of terms approaches infinity.
The geometric series, which serves as the foundation for many other series, is represented by the formula:

• \[ \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} \] for \(|x| < 1\).

This fundamental series leads to many interesting results when manipulated, such as shifting, differentiating, or multiplying by constants. The geometric series is convergent, meaning its sum exists, when \(|x| < 1\), which gives a clear criterion for convergence. Understanding these convergence properties is critical when dealing with infinite series.

Power Series

Power series provide a robust tool for expressing functions and analyzing their properties within certain intervals. Unlike a simple geometric series, a power series includes coefficients that influence each term. The general form of a power series is:

• \[ \sum_{n=0}^{\infty} a_n x^n \]

where \(a_n\) are coefficients, and \(x\) is the variable. Power series are used frequently in calculus because they can be differentiated and integrated term-by-term within their interval of convergence. The manipulation of a power series can yield significant information about a function over its radius of convergence.
In many advanced problems, like those related to differentiation within series, understanding the relationships between power series and other series types is key to solving complex equations.

Differentiation of Series

Differentiating a series is a powerful technique to transform and analyze series sums. For instance, taking the derivative of the geometric series, \(\sum_{n=0}^{\infty} x^n\), with respect to \(x\) helps derive important results:

• \[ \frac{d}{dx} \sum_{n=0}^{\infty} x^n = \sum_{n=1}^{\infty} n x^{n-1} \]

This result demonstrates how differentiation allows us to explore new series that might not be evident from the original form.
Using differentiation repeatedly on existing series reveals deeper insights into their characteristics and yields answers to complex problems more intuitively. In the context of the exercise, differentiation helps find new sums involving factors like \(n\) and \(n(n-1)\), demonstrating its applicability.

Binomial Theorem

The Binomial Theorem is a central concept in algebra that provides a way to expand expressions raised to a power. It gives insight into combinations and probability, generating functions, and simplification of polynomial expressions. The theorem states:

• \[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]

where \(\binom{n}{k}\) represents the binomial coefficient, which determines the number of ways to choose \(k\) elements from \(n\) elements.
This theorem not only simplifies calculations but also enables transformations of series. Applying binomial expansions can provide solutions, such as simplifying large polynomial expressions or finding sums in series like those encountered in the exercise. It offers a linkage between algebra and calculus, expanding our toolbox for problem-solving when approaching series-related questions.