Problem 36 - 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? $$ \sum_{n=1}^{\infty}(\sqrt{n+1}-\sqrt{n})(x-3)^{n} $$

Step-by-Step Solution

Answer

The radius is 0.5, it converges absolutely for \(2.5 < x < 3.5\), with no conditional convergence.

Step 1: Apply the Ratio Test

To find the radius of convergence for the series \(\sum_{n=1}^{\infty}(\sqrt{n+1}-\sqrt{n})(x-3)^{n}\), we apply the ratio test. This involves examining the limit \(lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\), where \(a_n = (\sqrt{n+1}-\sqrt{n})(x-3)^n\).

Step 2: Simplify the Ratio

Calculate the successive term ratio:\[\left| \frac{(\sqrt{n+2} - \sqrt{n+1})(x-3)^{n+1}}{(\sqrt{n+1} - \sqrt{n})(x-3)^n} \right| = |x-3| \cdot \frac{\sqrt{n+2} - \sqrt{n+1}}{\sqrt{n+1} - \sqrt{n}}.\]

Step 3: Evaluate the Limit

As \(n \to \infty\), notice that both denominators and numerators involving radicals tend towards twice the rate of increase of their root values. Use the approximation \(\sqrt{n+2} \approx \sqrt{n} + \frac{1}{\sqrt{n}}\) to simplify:\[\frac{\sqrt{n+2} - \sqrt{n+1}}{\sqrt{n+1} - \sqrt{n}} \approx \frac{\frac{1}{\sqrt{n}}}{\frac{1}{2\sqrt{n}}} = 2.\]Thus the limit is \(|x-3| \times 2\).

Step 4: Solve for Radius of Convergence

To assure the series converges, set \(|x-3| \cdot 2 < 1\) or \(|x-3| < \frac{1}{2}\). This implies the radius of convergence is \(\frac{1}{2}\).

Step 5: Determine Interval of Convergence

The translation \(x - 3\) suggests the center is at \(x = 3\). Thus, \(2 \times |x - 3| < 1\) indicates:\[ 2 \left(-\frac{1}{2} < x - 3 < \frac{1}{2}\right), \text{ simplifying to } 2.5 < x < 3.5.\]At endpoints, \(x = 2.5\) and \(x = 3.5\), further analysis (like the limit comparison) is needed to determine convergence, often they result in divergence.

Step 6: Check Absolute Convergence

Absolute convergence for the series occurs within the interval \(2.5 < x < 3.5\). The function \(\sqrt{n+1} - \sqrt{n}\) quickly tends to a zero magnitude with large \(n\), allowing the polynomial \((x-3)^n\)'s magnitude to sufficiently dominate in infinite analysis to show convergence.

Step 7: Checks for Conditional Convergence

Since series with the powder function conversations are divergent at the endpoint, conditional convergence does not occur at \(x = 2.5\) or \(x = 3.5\). Hence there are no values \(x\) for which the series converges conditionally.

Key Concepts

Ratio Test

The Ratio Test is a valuable tool for determining the radius of convergence of a series. When applied to a series like \( \sum_{n=1}^{\infty}(\sqrt{n+1}-\sqrt{n})(x-3)^{n} \), it helps to examine the behavior of the ratio of successive terms. The formula used in the Ratio Test is:

• Find \( a_n \), which is each term of the series. For this series, \( a_n = (\sqrt{n+1}-\sqrt{n})(x-3)^n \).
• Consider the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
• If the limit is less than 1, the series converges absolutely.
• If the limit is greater than 1 or infinite, the series diverges.

Applying this test to our series involves simplifying the ratio to see how \(|x-3|\) compares to \lim.This specific limit yielded \(|x-3| \times 2 < 1\), leading to a simplification that shows our radius of convergence.

Interval of Convergence

The interval of convergence can be derived once you know the radius of convergence. Given that the limit with the Ratio Test provided \(|x-3| < \frac{1}{2} \), we can determine that the series converges when \(x\) is within a distance of \(\frac{1}{2}\) from its center, \(x = 3\). This means testing within the range:

• \(2.5 < x < 3.5\)

The endpoints \(x = 2.5\) and \(x = 3.5\) need separate checks to determine convergence behavior.Often, these require additional tests such as comparison tests or specific function analysis. In this exercise, endpoint evaluations typically show divergence.

Absolute Convergence

Absolute convergence refers to the behavior of a series when its terms are replaced by their absolute values. If a series converges absolutely, it means \( \sum |a_n| \) converges. In this exercise, we identified that between the interval 2.5 and 3.5, as \( n \) grows, the influence of the \( \sqrt{n+1} - \sqrt{n} \) term diminishes fasterdue to the decrease towards zero. Thus, \( (x-3)^n \) becomes the controlling factor for convergence, ensuring absolute convergence within this whole range.

Conditional Convergence

Conditional convergence is less straightforward because it happens when a series converges but does not converge absolutely. In the case of our series, we examined it at the boundaries of the interval of convergence. For conditional convergence, it's important to check these endpoints:

• If neither endpoint results in a series maintaining convergence conditions without absolute convergence holding, then conditional convergence doesn't occur.
• This exercise concludes that since the series diverges at both \(x = 2.5\) and \(x = 3.5\), there are no values where just convergence (but not absolute) holds within the extended interval's endpoints.

Thus, this series displays no values of \(x\) for which conditional convergence is applicable.