Problem 40 - Thomas Calculus

Question

In Exercises \(37-40,\) find the series' radius of convergence. $$ \sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{n^{2}} x^{n} $$

Step-by-Step Solution

Answer

The radius of convergence is \(e\).

Step 1: Identify the Series Form

The series is given by \(\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right)^{n^{2}} x^{n}\). It is a power series in terms of \(x\).

Step 2: Apply the Root Test

To find the radius of convergence, we will use the Root Test:\[\lim_{n \to \infty} \sqrt[n]{\left|a_n\right|}\] where \(a_n = \left(\frac{n}{n+1}\right)^{n^2} x^n\). We need to evaluate:\[\lim_{n \to \infty} \left( \left( \frac{n}{n+1} \right)^{n^2} \right)^{1/n} \cdot |x|^{1/n} \]

Step 3: Simplify the Expression

For \(b_n = \left(\frac{n}{n+1}\right)^{n^{2}}\), evaluate:\[b_n^{1/n} = \left(\frac{n}{n+1}\right)^{n}\]This simplifies using:\[\lim_{n \to \infty} \left(\frac{n}{n+1}\right)^{n} = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right)^{n} = \lim_{n \to \infty} e^{-1} = \frac{1}{e}\]Therefore, we have:\[\lim_{n \to \infty} \sqrt[n]{b_n} = \frac{1}{e}\]

Step 4: Conclusion from Root Test

Since:\[\lim_{n \to \infty} \sqrt[n]{|a_n|} = \frac{1}{e}|x| \]The series converges when this is less than 1:\[\frac{1}{e}|x| < 1\]Solving gives the radius of convergence \(R\):\[|x| < e\]Thus, the radius of convergence is \(e\).

Key Concepts

Power Series

A power series is a series in the form:

• \( \sum_{n=0}^{\infty} a_n x^n \)

Here, each term of the series is built by multiplying a coefficient \(a_n\) with a power of \(x\). In the exercise, the series is:

• \( \sum_{n=1}^{\infty} \left(\frac{n}{n+1}\right)^{n^2} x^n \)

Here, the coefficients are given by \(a_n = \left(\frac{n}{n+1}\right)^{n^2}\) and the variable term by \(x^n\). This means the entire series changes as \(x\) changes. The convergence of the series depends on these changes, which is where the concept of radius of convergence comes in. The radius of convergence helps us understand the interval around \(x = 0\) where this series converges. Calculating this radius reveals when the series will result in a finite sum rather than diverging.

Root Test

The Root Test is an effective strategy to determine the convergence of a power series. The test assesses whether a series converges by examining the \(n\)-th root of the absolute value of its general term. For our exercise:

• We have \( a_n = \left( \frac{n}{n+1} \right)^{n^2} x^n \)

To apply the Root Test, calculate:

• \( \lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \left( \left( \frac{n}{n+1} \right)^{n^2} \right)^{1/n} \cdot |x|^{1/n} \)

Breaking it down, the \( b_n = \left(\frac{n}{n+1}\right)^{n^{2}} \) term simplifies using the property:

• \( \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right)^{n} = e^{-1} = \frac{1}{e} \)

This result allows us to conclude that if \( \frac{1}{e} |x| < 1 \), the series converges.

Limit Evaluation

To effectively apply the Root Test, a key step is to simplify the expression of \( \sqrt[n]{|a_n|} \) by evaluating limits. It's about understanding how each part of \(a_n\) behaves as \(n\) approaches infinity.For our series:

• First, simplify \( \left( \frac{n}{n+1} \right)^{n^2} \) and then take the \(n\)-th root, which reduces to evaluating \( \left( \frac{n}{n+1} \right)^n \)

This expression approximates:

• \( \left(1 - \frac{1}{n+1}\right)^n \) which approaches \( \frac{1}{e} \)

Using this in limit evaluation identifies the actual threshold for the variable \(|x|\):

• We need the limit \( \frac{1}{e} |x| < 1 \) to be satisfied for convergence.

Thus, solving \( |x| < e \) gives us the radius of convergence \(R = e\), indicating how far the series will converge around \(x = 0\).