Problem 5 - Thomas Calculus

Question

In Exercises \(1-36\) , (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely, (c) conditionally? $$ \sum_{n=0}^{\infty} \frac{(x-2)^{n}}{10^{n}} $$

Step-by-Step Solution

Answer

The series converges absolutely for \((-8, 12)\). It does not converge conditionally.

Step 1: Identify the given series

The given series is \( \sum_{n=0}^{\infty} \frac{(x-2)^{n}}{10^{n}} \). This is a power series centered at \( x = 2 \).

Step 2: Apply the ratio test

To find the radius of convergence, we apply the ratio test. Consider \( a_n = \frac{(x-2)^n}{10^n} \). The ratio test expression is given by \( \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x-2)^{n+1} / 10^{n+1}}{(x-2)^n / 10^n} \right| = \left| \frac{x-2}{10} \right| \).

Step 3: Set up inequality for convergence using the ratio test

The series converges when the absolute value of the ratio is less than 1: \( \left| \frac{x-2}{10} \right| < 1 \).

Step 4: Solve for the values of \(x\)

Solve the inequality \( \left| \frac{x-2}{10} \right| < 1 \) by considering \( -1 < \frac{x-2}{10} < 1 \). Multiply all parts of the inequality by 10 to get \( -10 < x-2 < 10 \).

Step 5: Determine the interval of convergence

Add 2 to each part of the inequality \( -10 < x-2 < 10 \) to find the interval of convergence: \( -8 < x < 12 \). Hence, the interval of convergence is \( (-8, 12) \).

Step 6: Determine absolute convergence

For absolute convergence, the series must converge when all terms are positive. In this series, it converges absolutely for the same interval \( (-8, 12) \) due to the ratio test covering absolute values.

Step 7: Check endpoints for conditional convergence

Check the endpoints \( x = -8 \) and \( x = 12 \). Substituting \( x = -8 \), the series becomes \( \sum_{n=0}^{\infty} \left( \frac{-10}{10} \right)^n = \sum_{n=0}^{\infty} (-1)^n \), which diverges. Substituting \( x = 12 \), the series becomes \( \sum_{n=0}^{\infty} 1^n \), which also diverges. Thus, there is no interval where it converges conditionally.

Key Concepts

Radius of Convergence

The radius of convergence is a crucial concept when dealing with power series. It tells us how far from a central point the series converges absolutely. For example, consider the power series given by \( \sum_{n=0}^{\infty} \frac{(x-2)^{n}}{10^{n}} \). By using the ratio test, which is a common method in finding the radius of convergence, we calculate the limit of the ratio of successive terms. In this case, it boils down to \( \left| \frac{x-2}{10} \right| \).

• Ratio Test: This method involves the limit \( \lim_{{n\to\infty}} \left| \frac{a_{n+1}}{a_n} \right| \).
• The series converges if the result of this ratio test is less than one.

Detecting that \( \left| \frac{x-2}{10} \right| < 1 \), we find that the expression restricts \( x \) to a distance of 10 from 2. Thus, the radius of convergence is 10. This means the series will converge absolutely for all \( x \) values within 10 units of the central point, excluding the boundaries.

Interval of Convergence

The interval of convergence takes the radius into account but also determines the actual values of \( x \) where the series converges. For our series, the absolute value inequality \( \left| \frac{x-2}{10} \right| < 1 \) transforms into \( -10 < x - 2 < 10 \). Solving this inequality, we adjust it to \( -8 < x < 12 \).

• This interval tells us that for any value of \( x \) between \(-8\) and \(12\), the series converges.
• The endpoints are not automatically included. We must verify them separately.

We tested \( x = -8 \) and \( x = 12 \) and found the series diverges at both. That means that the interval of convergence for this series is really \((-8, 12)\), not including the endpoints.

Absolute Convergence

Absolute convergence is when a series converges even if we replace all its terms with their absolute values. This is a stronger form of convergence. For the series \( \sum_{n=0}^{\infty} \frac{(x-2)^{n}}{10^{n}} \), absolute convergence occurs over the same range of \( x \) as regular convergence due to the nature of our ratio test result.

• In simple terms, if the series converges after making all terms positive, it absolutely converges.
• For this series, absolute convergence happens for \( x \) values within \(-8 < x < 12\).

When we apply the standard tests at endpoints, both failed to converge. As absolute convergence is not affected by the sign of terms, this confirms that it occurs only within the established interval, leaving no room for conditional convergence in this case.