Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\sum_{n=1}^{\infty} \frac{n^{2}+1}{3^{n}}\)

Step-by-Step Solution

Answer

The sum of the series is 2.

Step 1: Understanding the Series

The given problem is to evaluate the infinite series: \(\sum_{n=1}^{\infty} \frac{n^2 + 1}{3^n}\). This is an infinite series where the general term is \(\frac{n^2 + 1}{3^n}\).

Step 2: Break Down the General Term

The general term \(\frac{n^2+1}{3^n}\) can be split as \(\frac{n^2}{3^n} + \frac{1}{3^n}\). This allows us to separate the series into two simpler series: \(\sum_{n=1}^{\infty} \frac{n^2}{3^n}\) and \(\sum_{n=1}^{\infty} \frac{1}{3^n}\).

Step 3: Evaluate the Geometric Series

The series \(\sum_{n=1}^{\infty} \frac{1}{3^n}\) is a geometric series with the first term \(a = \frac{1}{3}\) and a common ratio \(r = \frac{1}{3}\). The sum of an infinite geometric series is given by \(\frac{a}{1-r}\). Thus, the sum is \(\frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}\).

Step 4: Transform the Non-Geometric Series

To evaluate \(\sum_{n=1}^{\infty} \frac{n^2}{3^n}\), we can use the formula \(\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}\) where \(|x| < 1\), and its related formulas for derivatives. Derivative-related formulas help in handling polynomial terms in the series.

Step 5: Apply Derivative to Geometric Series

Start with \(\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}\). The first derivative \(\sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2}\) by differentiating both sides. Multiply through by \(x\) to get \(\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}\).

Step 6: Derive Second Expression

Differentiate again to find \(\frac{d}{dx} \left( x \cdot \frac{x}{(1-x)^2} \right) = \sum_{n=1}^{\infty} n^2x^{n-1}\) leading to \(\sum_{n=1}^{\infty} n^2 x^n = \frac{x(1+x)}{(1-x)^3}\).

Step 7: Calculate \(\sum_{n=1}^{\infty} \frac{n^2}{3^n}\)

Set \(x = \frac{1}{3}\) in \(\sum_{n=1}^{\infty} n^2 x^n = \frac{x(1+x)}{(1-x)^3}\) to obtain \(\frac{\left(\frac{1}{3}\right)\left(1+\frac{1}{3}\right)}{\left(\frac{2}{3}\right)^3} = \frac{\frac{1}{3}\cdot\frac{4}{3}}{\frac{8}{27}} = \frac{\frac{4}{9}}{\frac{8}{27}} = \frac{4}{9} \times \frac{27}{8} = \frac{12}{8} = \frac{3}{2}\).

Step 8: Combine the Results

The sum of the entire series is the sum of the two parts found earlier: \(\sum_{n=1}^{\infty} \frac{n^2}{3^n} + \sum_{n=1}^{\infty} \frac{1}{3^n} = \frac{3}{2} + \frac{1}{2} = 2\).

Key Concepts

Geometric Series

A geometric series is a series where each term is a constant multiple, called the common ratio, of the previous term. The general form of a geometric series is given as:\[S = a + ar + ar^2 + ar^3 + \dots\]where \(a\) is the first term and \(r\) is the common ratio. In situations where the absolute value of \(r\) is less than 1, the series converges to a sum given by:\[S = \frac{a}{1-r}\]For example, in the problem above the series \(\sum_{n=1}^{\infty} \frac{1}{3^n}\) is a geometric series with \(a = \frac{1}{3}\) and \(r = \frac{1}{3}\). The sum of this series can be easily calculated as \(\frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}\).

Derivatives

Derivatives play a crucial role in evaluating series that are more complex than basic geometric series. In calculus, a derivative gives us the rate at which a function is changing at any given point. It is the fundamental concept behind finding slopes of curves.
By applying derivatives to the function \(f(x) = \frac{1}{1-x}\), such as taking the derivative of the geometric series formula, we can generate new series representations. For instance, taking the first derivative:\[\sum_{n=1}^{\infty} nx^{n} = \frac{x}{(1-x)^2}\]This step allows us to transform expressions involving polynomial terms. Derivatives help manipulate and solve series where the term involves powers of \(n\). This method is used to handle the non-geometric part of the original series, \(\sum_{n=1}^{\infty} \frac{n^2}{3^n}\).

Polynomial Terms

A polynomial term refers to expressions that involve powers of a variable. In our original problem, \(n^2+1\) is the polynomial embedded within the series.
Polynomials can make series more complex. In our task, we transform the polynomial in the numerator into manageable forms which are easier to evaluate. Breaking down \(\frac{n^2+1}{3^n}\) into \(\frac{n^2}{3^n} + \frac{1}{3^n}\) simplifies the series.
To evaluate sections of the series associated with polynomial terms, approaches like taking derivatives and transforming via known identities are commonly used. This process involves recognizing patterns in derivations and building solutions step-by-step by managing each segment separately.

Series Evaluation

Evaluating a series means finding the sum of all its terms, particularly for infinite series. The strategy usually involves simplifying and separating complex series into parts that are easier to handle.
The original problem required evaluating \(\sum_{n=1}^{\infty} \frac{n^2+1}{3^n}\). It was first broken down into two separate series: \(\sum_{n=1}^{\infty} \frac{n^2}{3^n}\) and \(\sum_{n=1}^{\infty} \frac{1}{3^n}\). Each of these parts is tackled using appropriate techniques—geometric series formula for one, and derivative manipulation for the other.
By carefully attacking each term, the final total is shown to be 2. Detailing and verifying each step within this problem-solving process ensures accuracy and reinforces the importance of methodical work when dealing with infinite series.