Problem 72 - Thomas Calculus

Question

In each of the geometric series in Exercises \(69-72,\) write out the first few terms of the series to find \(a\) and \(r,\) and find the sum of the series. Then express the inequality \(|r| < 1\) in terms of \(x\) and find the values of \(x\) for which the inequality holds and the series converges. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2}\left(\frac{1}{3+\sin x}\right)^{n} $$

Step-by-Step Solution

Answer

The series converges for all real numbers x.

Step 1: Identify the first term

The given series is \( \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2}\left(\frac{1}{3+\sin x}\right)^{n} \). The first term of a series, denoted \(a\), is the term when \(n=0\). Substituting \(n=0\), we have:\[a = \frac{(-1)^{0}}{2}\left(\frac{1}{3+\sin x}\right)^{0} = \frac{1}{2}.\]

Step 2: Determine the common ratio

The common ratio \(r\) of a geometric series is the factor that each term is multiplied by to obtain the subsequent term. In the given series, the common ratio is the entire expression raised to the \(n\)-th power since the powers differ by one between consecutive terms:\[r = \frac{1}{3+\sin x}.\]

Step 3: Find the sum of the series

The sum \(S\) of an infinite geometric series is given by the formula:\[S = \frac{a}{1-r},\]provided \(|r| < 1\). Here, we have:\[S = \frac{\frac{1}{2}}{1-\frac{1}{3+\sin x}} = \frac{1}{2} \cdot \frac{3+\sin x}{2+\sin x}.\]

Step 4: Set up the inequality |r| < 1

To find for which values of \(x\) the series converges, we need \(|r| < 1\). For the given common ratio \(r = \frac{1}{3+\sin x}\), this inequality becomes:\[\left| \frac{1}{3+\sin x} \right| < 1.\]

Step 5: Solve the inequality for x

Simplify the inequality:\[\left| \frac{1}{3+\sin x} \right| < 1 \implies 0 < \frac{1}{3+\sin x} < 1.\]Since \(\frac{1}{3+\sin x} > 0\) is always true when \(\sin x > -3\), we focus on the other side:\[\frac{1}{3+\sin x} < 1 \implies 1 < 3+\sin x \implies \sin x > -2.\]Therefore, the series converges when \(-2 < \sin x \leq 1\), which holds for all real \(x\).

Step 6: Conclusion: Values of x for Convergence

In conclusion, the series converges for all real numbers \(x\), as \(-2 < \sin x \leq 1\) for all real \(x\).

Key Concepts

Common Ratio

In a geometric series, the common ratio, denoted as \( r \), plays a crucial role. It is the constant factor between any two consecutive terms of the series. Understanding the common ratio is essential because it helps determine the behavior of the series.
In the given series \( \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2}\left(\frac{1}{3+\sin x}\right)^{n} \), the common ratio is identified by examining how each term is related to its predecessor. Here, the common ratio \( r \) is calculated as \( \frac{1}{3+\sin x} \).

• Start with the expression that changes with each term, which is the part of the series raised to the power \( n \).
• In this series, the sequence inside the parentheses raised to a power is \( \left(\frac{1}{3+\sin x}\right)^{n} \).
• The common ratio is, therefore, the expression \( \frac{1}{3+\sin x} \).

This ratio is fundamental in calculating further properties of the series, such as its sum.

Sum of Series

The sum of an infinite geometric series can provide significant insights into the series' behavior. For a geometric series with a first term \( a \) and a common ratio \( r \), the sum \( S \) of the series is given by the formula:\[ S = \frac{a}{1-r} \]provided that the absolute value of the common ratio, \(|r|\), is less than 1.
In the exercise, the first term \( a \) is \( \frac{1}{2} \), and the common ratio was found to be \( \frac{1}{3+\sin x} \). Therefore, the sum of the series is calculated as follows:

• Using the sum formula: \( S = \frac{\frac{1}{2}}{1 - \frac{1}{3+\sin x}} \).
• This simplifies to: \( S = \frac{1}{2} \cdot \frac{3+\sin x}{2+\sin x} \).

This result is only valid if the series converges, which is determined by the next concept.

Inequality for Convergence

Convergence in a geometric series is crucial for understanding when a series can actually sum to a finite value. The condition for convergence is tied to the common ratio's absolute value. The series converges only if \( |r| < 1 \).
For the discussed series, with \( r = \frac{1}{3+\sin x} \), the convergence inequality becomes \( \left| \frac{1}{3+\sin x} \right| < 1 \).

• This inequality means the series will converge if the expression \( \frac{1}{3+\sin x} \) is between -1 and 1.
• Since \( \frac{1}{3+\sin x} > 0 \) when \( 3+\sin x > 0 \), we simplify the condition to focus on:
  \( 0 < \frac{1}{3+\sin x} < 1 \).
• This can further be simplified to \( 1 < 3+\sin x \), indicating \( \sin x > -2 \).

Since \( \sin x \) oscillates between -1 and 1 for all real \( x \), the condition \( -2 < \sin x \leq 1 \) is always satisfied. Thus, the series converges for all real numbers \( x \), ensuring the finite sum is valid.