Question

Use the root and limit comparison tests to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges. $$ \left.a_{n}=1 / x_{n}^{n} \text { where } x_{n+1}=\frac{1}{2} x_{n}+\frac{1}{x_{n}}, x_{1}=1 \text { (Hint: Find limit of }\left\\{x_{n}\right\\} .\right) $$

Short answer

The series \( \sum_{n=1}^{\infty} a_n \) converges.

Step-by-step solution

Analyze the Sequence x_n

Given that \( x_{n+1} = \frac{1}{2}x_{n} + \frac{1}{x_{n}} \) with \( x_1 = 1 \), let's analyze this recurrence relation. Over time, \( x_n \) converges to a fixed point where \( x_{n+1} = x_n = x \). Set \( x = \frac{1}{2}x + \frac{1}{x} \) and solve for \( x \).

Solve for the Fixed Point of x_n

Substitute \( x \) into the fixed-point equation: \[ x = \frac{1}{2}x + \frac{1}{x}. \] Rearrange to get \( x - \frac{1}{2}x = \frac{1}{x} \), leading to \( \frac{1}{2}x = \frac{1}{x} \). Thus, \( x^2 = 2 \), meaning \( x = \sqrt{2} \). So, \( x_n \to \sqrt{2} \) as \( n \to \infty \).

Define and Analyze a_n

For \( a_n = \frac{1}{x_n^n} \), and with the limit \( x_n \to \sqrt{2} \), determine the behavior of \( a_n \) as \( n \to \infty \). The sequence becomes \( a_n \approx \frac{1}{(\sqrt{2})^n} \), which behaves like \( (\frac{1}{\sqrt{2}})^n \).

Apply Root Test

The root test is given by \( \lim_{n \to \infty} \sqrt[n]{a_n} = \rho \). Substitute \( a_n = \frac{1}{x_n^n} \) to get \( \lim_{n \to \infty} \sqrt[n]{\frac{1}{(\sqrt{2})^n}} = \frac{1}{\sqrt{2}} \). Since \( \frac{1}{\sqrt{2}} < 1 \), the series converges.

Apply Limit Comparison Test

Next, use the limit comparison test with \( b_n = (\frac{1}{\sqrt{2}})^n \). We have \( \lim_{n \to \infty} \frac{a_n}{b_n} = 1 \), which is a finite nonzero number. As \( \sum b_n \) converges (it's a geometric series with ratio \( \frac{1}{\sqrt{2}} < 1 \)), by the limit comparison test, \( \sum a_n \) converges as well.

Key concepts

Root Test

The root test is a valuable tool for determining the convergence of an infinite series. It works by examining the behavior of the sequence elements as they grow. Specifically, the test considers the n-th root of the terms of the series.
For a given series \( \sum_{n=1}^{\infty} a_n \), we compute\[ \rho = \lim_{n \to \infty} \sqrt[n]{|a_n|} \]After finding \( \rho \), we can make conclusions as follows:

• If \( \rho < 1 \), the series converges absolutely.
• If \( \rho > 1 \) or \( \rho = \infty \), the series diverges.
• If \( \rho = 1 \), the test is inconclusive.

In this exercise, applying the root test led to \( \rho = \frac{1}{\sqrt{2}} \). Since \( \frac{1}{\sqrt{2}} < 1 \), it indicates that the series \( \sum_{n=1}^{\infty} a_n \) converges.

Limit Comparison Test

Another powerful method to decide the convergence of a series is the limit comparison test. It uses a comparison between the series of interest and a reference series that is known to converge or diverge.
For series \( \sum_{n=1}^{\infty} a_n \) and \( \sum_{n=1}^{\infty} b_n \), the limit comparison test examines:\[ \lim_{n \to \infty} \frac{a_n}{b_n} = L \]If \( L \) is a positive finite number, both series either converge or diverge together.
In our exercise, we compared \( a_n = \frac{1}{x_n^n} \) to \( b_n = \left(\frac{1}{\sqrt{2}}\right)^n \). The limit was found to be 1, confirming convergence of both series because \( \sum_{n=1}^{\infty} b_n \) is a convergent geometric series.

Series Convergence

Understanding series convergence is crucial in mathematics since it determines whether an infinite sum reaches a finite value. Different tests can verify convergence, such as the root test and limit comparison test highlighted herein.
A series \( \sum_{n=1}^{\infty} a_n \) converges if its partial sums tend towards a finite number as \( n \to \infty \). Often, mathematical tools are necessary to find whether this happens.
The exercise involved showed convergent behavior through calculated tests, using both root test and limit comparison lending confidence to the conclusion. Recognizing these patterns makes predicting convergence or divergence more intuitive with practice.

Geometric Series

Geometric series are a specific type of series where each term is a constant multiple of the previous term. They are of the form:\[ \sum_{n=1}^{\infty} ar^n \]where \( a \) is the first term and \( r \) is the common ratio.

• If the absolute value of \( r < 1 \), the geometric series converges.
• If \( |r| \geq 1 \), the series diverges.

In this exercise, the series \( \sum_{n=1}^{\infty} \left(\frac{1}{\sqrt{2}}\right)^n \) is a geometric series with \( a = 1 \) and \( r = \frac{1}{\sqrt{2}} \). With \( r < 1 \), it converges. This provided a critical comparison point in applying the limit comparison test, helping confirm the convergence of the original series.