Problem 18 - Calculus: Early Transcendentals

Question

Find the radius of convergence and interval of convergence of the series. \( \sum_{n = 1}^{\infty} \frac {\sqrt{n}}{8^n} (x + 6)^n \)

Step-by-Step Solution

Answer

Radius is 8; interval is \((-14, 2)\).

Step 1: Identify the Series Details

The given series is \[\sum_{n=1}^{\infty} a_n (x+6)^n \]where \( a_n = \frac{\sqrt{n}}{8^n} \). We aim to find the radius of convergence and interval of convergence of this power series.

Step 2: Apply the Ratio Test

To find the radius of convergence, apply the ratio test to \( a_n (x+6)^n \). Compute\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} (x+6) \right|.\]Substituting \( a_n = \frac{\sqrt{n}}{8^n} \),\[\left| \frac{a_{n+1}}{a_n} \right| = \frac{\sqrt{n+1}}{8^{n+1}} \cdot \frac{8^n}{\sqrt{n}} = \frac{\sqrt{n+1}}{\sqrt{n} \cdot 8}.\]Thus,\[\left| \frac{a_{n+1}(x+6)^{n+1}}{a_n(x+6)^n} \right| = \left| \frac{\sqrt{n+1}}{\sqrt{n} \cdot 8} \cdot (x+6) \right|.\]

Step 3: Simplify the Limit Expression

The limit from the Ratio Test becomes:\[\lim_{n \to \infty} \left| \frac{\sqrt{n+1}}{\sqrt{n} \cdot 8} (x+6) \right| = \frac{|x+6|}{8} \cdot \lim_{n \to \infty} \sqrt{\frac{n+1}{n}}.\]Knowing that \( \lim_{n \to \infty} \sqrt{\frac{n+1}{n}} = 1 \), the limit simplifies to\[\frac{|x+6|}{8}.\]

Step 4: Apply the Convergence Condition

For the series to converge, the result of the Ratio Test must be less than 1:\[\frac{|x+6|}{8} < 1.\]Solving this inequality gives\[|x+6| < 8.\]This implies that\[-8 < x+6 < 8.\]

Step 5: Simplify the Interval

Rewriting the inequality\[-8 < x+6 < 8\]leads to\[-14 < x < 2.\]The radius of convergence \( R \) is 8, derived from the inequality \( |x+6| < 8 \).

Step 6: Check Endpoints for Convergence

Plug in \( x = -14 \) and \( x = 2 \) into the original series to check for convergence at the endpoints.- For \( x = -14 \), the series becomes: \[ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{8^n} (-8)^n, \] which is a divergent series due to the form \( (-1)^n n^{1/2} \).- For \( x = 2 \), the series becomes: \[ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{8^n} 8^n = \sum_{n=1}^{\infty} \sqrt{n}, \] which also diverges.

Key Concepts

Interval of Convergence

The interval of convergence of a power series is the range of values of the variable for which the series converges. In this exercise, we start by setting up the problem: a series \( \sum_{n=1}^{\infty} \frac{\sqrt{n}}{8^n}(x + 6)^n \). Our goal is to determine where this series converges around a central point, which is \( x = -6 \) in this case.

To find this interval, we usually solve the inequality obtained from applying the Ratio Test. For example, from \( \frac{|x+6|}{8} < 1 \), we solve for \( x \), which gives us \(-14 < x < 2 \). This tells us that the series converges for any \( x \) between \(-14 \) and \( 2 \), not including these endpoints initially.

Endpoints require special attention as they can sometimes be part of the interval. You need to test them separately to see if the series converges at these points, which we will look into further.

Ratio Test

The Ratio Test is a powerful tool for determining the convergence of a series. It involves looking at the limit of the absolute value of the ratio of consecutive terms, specifically \( \lim_{n \to \infty} \left| \frac{a_{n+1}(x+6)^{n+1}}{a_n(x+6)^n} \right| \).

In this problem, the Ratio Test helps find the radius of convergence. We compute the limit: \( \lim_{n \to \infty} \left| \frac{\sqrt{n+1}/8 \cdot (x+6)}{\sqrt{n}} \right| = \frac{|x+6|}{8} \).

The series converges if this result is less than 1. Thus, setting \( \frac{|x+6|}{8} < 1 \) provides the necessary condition for convergence. Solving this inequality gives the limits of \( x \) that keep the series converging, leading to the interval discussed previously.

Power Series

A power series is a series of the form \( \sum_{n=0}^{\infty} a_n (x-c)^n \), centered at a point \( c \). Each \( a_n \) represents coefficients which, together with \( x \), determine the behavior of the series.

This exercise deals with a particular power series where \( a_n = \frac{\sqrt{n}}{8^n} \) and \( c = -6 \). Such series are unique in their utility because they can converge to a specific function in a neighborhood on the number line.

The concept of convergence and radius is critical: **the radius of convergence** shows how far from \( c \) the series still represents the function faithfully. Understanding this helps solve more complex functions and their behaviors over different intervals.

Convergence Test

A convergence test checks whether a series converges or diverges. Many tests exist, each suitable for different types of series. In this exercise, the Ratio Test was chosen because it aligns well with pinpointing the radius of convergence for power series.

After using the Ratio Test and finding the interval of convergence \((-14, 2) \), checking **endpoint convergence** is essential, however. For \( x = -14 \), the term simplifies and diverges. Similarly, for \( x = 2 \), the series remains divergent.

Engaging with convergence tests provides clarity about which tests serve specific functions best, validating whether series reach convergence or not, which is crucial when dealing with functions within certain limits.