/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Find the interval of convergence... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the interval of convergence of each power series. \(\sum_{n=1}^{\infty} \frac{n^{10} x^{n}}{10^{n}}\)

Short Answer

Expert verified
The interval of convergence is \(-10 < x < 10\).

Step by step solution

01

Identify the Power Series

The given power series is \( \sum_{n=1}^{\infty} \frac{n^{10} x^{n}}{10^{n}} \).Let's find the interval of convergence by applying the Ratio Test.
02

Apply the Ratio Test

The Ratio Test requires us to consider the limit:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]For our series, the terms are \( a_n = \frac{n^{10} x^{n}}{10^{n}} \). So,\[a_{n+1} = \frac{(n+1)^{10} x^{n+1}}{10^{n+1}} = \frac{(n+1)^{10} x^{n} x}{10^{n+1}}\]Forming the ratio:\[\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1)^{10} x}{10 \cdot n^{10}} \right|\]
03

Compute the Limit using Ratio Test

Compute the limit:\[\lim_{n \to \infty} \left| \frac{(n+1)^{10} x}{10 n^{10}} \right| = \left| x \right| \cdot \lim_{n \to \infty} \frac{(n+1)^{10}}{10 n^{10}}\]Simplifying the limit, we see: \[= \left| x \right| \cdot \frac{1}{10}\]Thus, the limit becomes \( \frac{\left| x \right|}{10} \). The Ratio Test tells us the series converges when this limit is less than 1.
04

Determine the Radius and Interval of Convergence

Setting the limit less than 1:\[\frac{\left| x \right|}{10} < 1\]Simplifying, we get:\[\left| x \right| < 10\]Thus, the radius of convergence is 10. The series converges on the interval \(-10 < x < 10\).
05

Check Endpoints for Convergence

To find the interval of convergence, we must test the convergence at the endpoints \( x = -10 \) and \( x = 10 \).1. **For \( x = 10 \):** The series becomes \( \sum_{n=1}^{\infty} \frac{n^{10} 10^{n}}{10^{n}} = \sum_{n=1}^{\infty} n^{10} \), which diverges.2. **For \( x = -10 \):** The series becomes \( \sum_{n=1}^{\infty} \frac{n^{10} (-10)^{n}}{10^{n}} = \sum_{n=1}^{\infty} (-1)^n n^{10} \), which also diverges.Therefore, the series diverges at both endpoints.

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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 an infinite series of the form \( \sum_{n=0}^{\infty} a_n (x - c)^n \), where \( a_n \) represents the sequence of coefficients and \( c \) is the center of the series. In the series given, \( \sum_{n=1}^{\infty} \frac{n^{10} x^{n}}{10^{n}} \), the power series is centered around zero, making it a straightforward case without shifts. The series involves powers of \( x \) and is characterized by the coefficients \( n^{10}/10^n \).

These types of series are fundamental in approximating functions and appear frequently in calculus. They give us a way to express complex functions as infinite polynomial-like series, which can be more manageable. Understanding how power series function is crucial for approximating functions and understanding convergence.
Ratio Test
The Ratio Test is a powerful method to determine the convergence of series, especially useful for power series. The Ratio Test examines the limit of the absolute value of the ratio of successive terms in a series. For a series \( \sum a_n \), the test involves computing:

\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]

If the limit is less than one, the series converges absolutely. If it is more than one, the series diverges. If it equals one, the test is inconclusive.

In our exercise, we applied the Ratio Test to the series \( a_n = \frac{n^{10} x^{n}}{10^{n}} \). The process involved simplifying the ratio \( \left| \frac{a_{n+1}}{a_n} \right| \) and evaluating it as \( n \to \infty \). With the result \( \lim_{n \to \infty} \frac{(n+1)^{10} x}{10 n^{10}} = \frac{|x|}{10} \), the series converges for \( \frac{|x|}{10} < 1 \), showing the importance of simplifying ratios for convergence.
Radius of Convergence
The radius of convergence is a key concept in understanding where a power series converges. It's the distance from the center of convergence \( c \) within which all the terms of the series sum up to a finite number.

For our series, we determined that \( \frac{|x|}{10} < 1 \), which simplifies to \( |x| < 10 \). This indicates that the radius of convergence is 10, meaning that the series converges when \( x \) is within 10 units of the center, zero, thus \(-10 < x < 10\).

Knowing the radius helps in analyzing the behavior of the series, especially when trying to approximate functions. Within this radius, the series behaves nicely and sums to a finite value. Beyond the radius, the convergence is not guaranteed.
Endpoints Check
Checking the endpoints of the interval of convergence is essential because the series might behave differently at these points. While the Ratio Test gives a general interval \(-10 < x < 10\), the convergence at \( x = -10 \) and \( x = 10 \) must be verified separately.

For \( x = 10 \), the series transforms to \( \sum_{n=1}^{\infty} n^{10} \), which is known to diverge because the terms do not approach zero. Similarly, for \( x = -10 \), the series becomes \( \sum_{n=1}^{\infty} (-1)^n n^{10} \), which also diverges.

These checks are crucial because they ensure that our understanding of the interval does not leave out any convergent parts at the boundaries. Endpoint checks confirm that in this problem, the series does not converge at the endpoints \( -10 \) and \( 10 \), leaving the interval of convergence as \( -10 < x < 10 \).

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Most popular questions from this chapter

Use power series established in this section to find a power series representation of the given function. Then determine the radius of convergence of the resulting series. \(f(x)=\cos ^{2} 2 x=\frac{1}{2}(1+\cos 4 x)\)

Use an infinite series to approximate the indicated number accurate to three decimal places. \(\cos \left(\frac{\pi}{20}\right)\)

Suppose that \(\sum a_{n}\) is a conditionally convergent infinite series. For each \(n\), let $$a_{n}^{+}=\frac{a_{n}+\left|a_{n}\right|}{2} \text { and } a_{n}=\frac{a_{n}-\left|a_{n}\right|}{2} \text { . }$$ (a) Explain why \(\Sigma a_{n}^{+}\) consists of the positive terms of \(\Sigma a_{n}\) and why \(\Sigma a_{n}^{-}\) consists of the negative terms of \(\suma_{n}\). (b) Given a real number \(r\), show that some rearrangement of the conditionally convergent series \(\Sigma a_{n}\) converges to \(r\). (Snggestion: If \(r\) is positive, for instance, begin with the first partial sum of the positive series \(\Sigma a_{n}^{+}\) that exceeds \(r\). Then add just enough terms of the negative series \(\Sigma a_{n}\) so that the cumulative sum is less than \(r\). Next add just enough terms of the positive series that the cumulative sum is greater than \(r\), and continuc in this way to define the desired rearrangement.) Why does it follow that this rearranged infinite series converges to \(r\) ?

Find a power series representation for the given function \(f(x)\) by using termwise integration. \(f(x)=\int_{0}^{x} \frac{\arctan t}{t} d t\)

(a) Suppose that \(r\) is a (fixed) number such that \(|r|<1\). Use the ratio test to prove that the series \(\sum_{n=0}^{2} n r^{n}\) con- verges. Let \(S\) denote its sum. (b) Show that $$(1-r) S=\sum_{n=1}^{\infty} r^{n} .$$ Show how to conclude that $$\sum_{n=0}^{\infty} n r^{n}=\frac{r}{(1-r)^{2}}$$

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