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Chapter 10: Problem 40

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

Short Answer

Expert verified
The radius of convergence is \(e\).

Step by step solution

01

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 \(x\).
02

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:\[\lim_{n \to \infty} \left( \left( \frac{n}{n+1} \right)^{n^2} \right)^{1/n} \cdot |x|^{1/n} \]
03

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:\[\lim_{n \to \infty} \sqrt[n]{b_n} = \frac{1}{e}\]
04

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 \(e\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \(x = 0\).

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Most popular questions from this chapter

Quadratic Approximations The Taylor polynomial of order 2 generated by a twice-differentiable function \(f(x)\) at \(x=a\) is called the quadratic approximation of \(f\) at \(x=a.\) find the (a) linearization (Taylor polynomial of order 1) and (b) quadratic approximation of \(f\) at \(x=0\). \(f(x)=1 / \sqrt{1-x^{2}}\)

Use series to evaluate the limits. \begin{equation} \lim _{x \rightarrow 0} \frac{\ln \left(1+x^{2}\right)}{1-\cos x} \end{equation}

Quadratic Approximations The Taylor polynomial of order 2 generated by a twice-differentiable function \(f(x)\) at \(x=a\) is called the quadratic approximation of \(f\) at \(x=a.\) find the (a) linearization (Taylor polynomial of order 1) and (b) quadratic approximation of \(f\) at \(x=0\). \(f(x)=\cosh x\)

Show that if \(\sum_{n=1}^{\infty} a_{n}\) and \(\sum_{n=1}^{\infty} b_{n}\) both converge absolutely, then so do the following. $$\begin{array}{ll}{\text { a. }} & {\sum_{n=1}^{\infty}\left(a_{n}+b_{n}\right)} & {\text { b. } \sum_{n=1}^{\infty}\left(a_{n}-b_{n}\right)} \\ {\text { c. }} & {\sum_{n=1}^{\infty} k a_{n}(k \text { any number })}\end{array}$$

Approximation properties of Taylor polynomials Suppose that \(f(x)\) is differentiable on an interval centered at \(x=a\) and that \(g(x)=b_{0}+b_{1}(x-a)+\cdots+b_{n}(x-a)^{n}\) is a polynomial of degree \(n\) with constant coefficients \(b_{0}, \ldots, b_{n}\) . Let \(E(x)=\) \(f(x)-g(x) .\) Show that if we impose on \(g\) the conditions i) \(E(a)=0\) ii) $$\lim _{x \rightarrow a} \frac{E(x)}{(x-a)^{n}}=0$$ then $$\begin{array}{r}{g(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots} \\ {+\frac{f^{(n)}(a)}{n !}(x-a)^{n}}\end{array}$$ Thus, the Taylor polynomial \(P_{n}(x)\) is the only polynomial of degree less than or equal to \(n\) whose error is both zero at \(x=a\) and negligible when compared with \((x-a)^{n}.\)
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