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Magazine Chapter 11: Problem 35
Determine the radius and interval of convergence of the following power series. $$\sum_{k=0}^{\infty} \frac{k^{20} x^{k}}{(2 k+1) !}$$
Short Answer
Expert verified
Answer: The radius of convergence is infinity, and the interval of convergence is $$(-\infty, +\infty)$$.
Step by step solution
01
Applying the Ratio Test
To find the radius of convergence, apply the Ratio Test by calculating the limit:
$$\lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| = L$$
where $$a_k = \frac{k^{20} x^{k}}{(2 k+1) !}$$
02
Ratio of consecutive terms
Find the ratio of consecutive terms:
$$\frac{a_{k+1}}{a_k} = \frac{\frac{(k+1)^{20} x^{k+1}}{(2(k+1)+1) !}}{\frac{k^{20} x^{k}}{(2 k+1) !}}$$
By simplifying, we get:
$$\frac{a_{k+1}}{a_k} = \frac{(k+1)^{20}}{k^{20}} \cdot \frac{x^{k+1}}{x^{k}} \cdot \frac{(2k+1)!}{(2k+3)!}$$
03
Simplify the ratio
Further simplifying the equation:
$$\frac{a_{k+1}}{a_k} = \left(\frac{k+1}{k}\right)^{20} \cdot x \cdot \frac{1}{(2k+2)(2k+3)}$$
04
Compute the limit
Next, compute the limit as k approaches infinity:
$$\lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| = \lim_{k \to \infty} \left|\left(\frac{k+1}{k}\right)^{20} \cdot x \cdot \frac{1}{(2k+2)(2k+3)}\right|$$
This limit can be simplified as:
$$= |x| \lim_{k \to \infty} \frac{1}{(2k+2)(2k+3)}$$
The limit of the last expression is 0, so our limit becomes:
$$= 0 \cdot |x| = 0$$
05
Determine the radius of convergence
For the series to converge, the ratio test limit L should be less than 1. In our case, L = 0, which is less than 1. Therefore, the series converges for all x. The radius of convergence (R) is thus infinite, which means it converges for all values of x:
$$R = \infty$$
06
Determine the interval of convergence
Since the radius of convergence is infinite, the interval of convergence is:
$$(-\infty, +\infty)$$
So, the radius of convergence is infinity, and the interval of convergence is $$(-\infty, +\infty)$$.
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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 sum of the form \( \sum_{n=0}^{\infty} a_n (x - c)^n \), where \( a_n \) represents the coefficient of the nth term, \( x \) is the variable, and \( c \) is the center of the series. Each power series has a radius of convergence around this center point within which the series converges to a well-defined value.
Understanding power series is critical in fields such as mathematics, engineering, and physics, as they are used to represent functions in a form that is easy to manipulate. The series you are trying to find the convergence for is one such example, except the center here is zero, making it a Maclaurin series. Power series can be used to approximate complex functions using polynomials to make calculations more manageable.
Understanding power series is critical in fields such as mathematics, engineering, and physics, as they are used to represent functions in a form that is easy to manipulate. The series you are trying to find the convergence for is one such example, except the center here is zero, making it a Maclaurin series. Power series can be used to approximate complex functions using polynomials to make calculations more manageable.
Ratio Test
The Ratio Test is a method used to determine the absolute convergence of an infinite series. To apply the test, you consider the limit of the absolute value of the ratio of consecutive terms \( \lim_{k \to \infty} |a_{k+1} / a_k| = L \). If this limit \( L \) is less than 1, the series converges absolutely; if \( L \) is greater than 1, the series diverges; and if \( L \) equals 1, the test is inconclusive.
Applying the Ratio Test
In the solution you're studying, the Ratio Test simplifies to finding the limit of the expression \( |x| \lim_{k \to \infty} 1 / ((2k+2)(2k+3)) \), which tends to zero. Since the limit is less than 1, the series converges. However, when this method yields zero, the implication is profound - it suggests that convergence occurs for all real numbers \( x \) since multiplying zero by any \( x \) results in zero, thus satisfying the condition \( L < 1 \) for all \( x \) in the real number system.Interval of Convergence
The interval of convergence for a power series is the set of all real numbers \( x \) for which the series converges. It is closely related to the radius of convergence \( R \); specifically, the interval of convergence is \( (c - R, c + R) \), where \( c \) is the center of the series. If \( R \) is infinite, the interval of convergence is the entire set of real numbers, \( (-\infty, +\infty) \).
In your exercise, since the radius of convergence is infinite, the series converges for all \( x \) without any restrictions, signifying that the interval of convergence is indeed \( (-\infty, +\infty) \) as determined in Step 6 of the solution. This is a special case since typically the radius of convergence is a finite number, resulting in a bounded interval of convergence. An infinite interval indicates that the function defined by the power series is entire, meaning it is defined and analytic for every complex number.
In your exercise, since the radius of convergence is infinite, the series converges for all \( x \) without any restrictions, signifying that the interval of convergence is indeed \( (-\infty, +\infty) \) as determined in Step 6 of the solution. This is a special case since typically the radius of convergence is a finite number, resulting in a bounded interval of convergence. An infinite interval indicates that the function defined by the power series is entire, meaning it is defined and analytic for every complex number.
