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Magazine Chapter 11: Problem 9
Find the radius of convergence and interval of convergence of the series. $$ \sum_{n=1}^{\infty} \frac{x^{n}}{n^{4} 4^{n}} $$
Short Answer
Expert verified
Radius of convergence is 4; interval of convergence is \([-4, 4]\).
Step by step solution
01
Identify the Series Form
The given series is \( \sum_{n=1}^{\infty} \frac{x^{n}}{n^{4} 4^{n}} \). This is a power series of the form \( \sum_{n=0}^{\infty} c_n x^n \) with \( c_n = \frac{1}{n^{4} 4^{n}} \), and it's centered at 0.
02
Use the Ratio Test
To find the radius of convergence, we use the Ratio Test. Consider the ratio \( \left| \frac{c_{n+1} x^{n+1}}{c_n x^n} \right| \). For our series, this becomes:\[\frac{x^{n+1}}{(n+1)^4 4^{n+1}} \cdot \frac{n^4 4^n}{x^n} = \left| x \right| \cdot \frac{n^4}{(n+1)^4} \cdot \frac{1}{4}\]
03
Simplify the Ratio
Simplify the expression from Step 2:\[\left| \frac{x n^4}{4 (n+1)^4} \right| \] As \( n \to \infty \), \( \frac{n^4}{(n+1)^4} \to 1 \), so the limit of the ratio is \( \left| \frac{x}{4} \right| \).
04
Apply the Ratio Test
For the series to converge, the limit of the ratio must be less than 1.\[\left| \frac{x}{4} \right| < 1\] This implies \( \left| x \right| < 4 \). The radius of convergence is 4.
05
Find the Interval of Convergence
Given the radius of convergence is 4, the interval is \( -4 < x < 4 \). We need to check the endpoints separately by substituting \( x = -4 \) and \( x = 4 \) into the series to determine the convergence at the endpoints.
06
Check Endpoint \( x = 4 \)
Substitute \( x = 4 \) into the series:\[\sum_{n=1}^{\infty} \frac{4^n}{n^4 4^n} = \sum_{n=1}^{\infty} \frac{1}{n^4}\] This is a \( p \)-series with \( p = 4 > 1 \), which converges.
07
Check Endpoint \( x = -4 \)
Substitute \( x = -4 \) into the series:\[\sum_{n=1}^{\infty} \frac{(-4)^n}{n^4 4^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^4}\] This series is an alternating \( p \)-series with \( p = 4 > 1 \), which also converges by the Alternating Series Test.
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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 particular type of series that can be thought of as an infinite polynomial. It has the general form:
- \( \sum_{n=0}^{\infty} c_n (x - a)^n \)
- where \( c_n \) are the coefficients, \( x \) is the variable, and \( a \) is the center of the series.
- Each term in this series is of the form \( \frac{x^n}{n^4 \cdot 4^n} \).
- It's centered at \( x = 0 \).
Ratio Test
The Ratio Test is a method to determine the convergence or divergence of a series. In essence, it involves calculating the limit of the ratio of consecutive terms:
- Consider a series \( \sum_{n=0}^{\infty} a_n \).
- Calculate \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
- If this limit is less than 1, the series converges absolutely. If it is greater than 1, the series diverges. If it equals 1, the test is inconclusive.
- Here we consider \( c_n = \frac{1}{n^4 \cdot 4^n} \).
- We simplify the ratio to \( \left| \frac{x n^4}{4 (n+1)^4} \right| \) and find its limit as \( n \to \infty \), which is \( \left| \frac{x}{4} \right| \).
- \( \left| \frac{x}{4} \right| < 1 \).
- This gives the radius of convergence as 4.
Interval of Convergence
The interval of convergence defines the set of \( x \) values for which a power series converges. To find this, we use the radius of convergence:
- If the radius of convergence is \( R \), the interval is \( (a-R, a+R) \), where \( a \) is the center of the series.
- It's centered at 0 and has a radius of 4. Hence, the interval of convergence is \( -4 < x < 4 \).
- For \( x = 4 \), the series becomes a \( p \)-series with \( p = 4 \), which converges.
- For \( x = -4 \), the series remains an alternating \( p \)-series with \( p = 4 \), also converging by the alternating series test.
P-Series
A \( p \)-series is a specific type of series with the form:
- \( \sum_{n=1}^{\infty} \frac{1}{n^p} \).
- If \( p > 1 \), the series converges.
- If \( p \leq 1 \), it diverges.
- For \( x = 4 \), we have \( \sum_{n=1}^{\infty} \frac{1}{n^4} \), which is a convergent \( p \)-series since \( p = 4 \).
- For \( x = -4 \), the series forms an alternating \( p \)-series: \( \sum_{n=1}^{\infty} (-1)^n \frac{1}{n^4} \), also converging thanks to the alternating series test.
