Radius And Interval Of Convergence For Power Series Explained

This article delves into the crucial topic of determining the radius and interval of convergence for various power series. Understanding these concepts is fundamental in the study of infinite series and their applications in calculus, differential equations, and complex analysis. We will explore several examples, applying the ratio test and other techniques to rigorously find the radius and interval of convergence for each series.

(a) ${\sum_{n=1}^{\infty} \frac{(x+4)^n}{n(3^n)}}$

To determine the radius and interval of convergence for the power series ${\sum_{n=1}^{\infty} \frac{(x+4)^n}{n(3^n)}}$, we will employ the ratio test. The ratio test is a powerful tool for assessing the convergence of infinite series, especially those involving factorials or exponential terms. It involves examining the limit of the ratio of consecutive terms in the series. Let's define the general term of the series as:

${ a_n = \frac{(x+4)^n}{n(3^n)} }$

Now, we compute the ratio of the (n+1)-th term to the n-th term:

${ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{(x+4)^{n+1}}{(n+1)(3^{n+1})}}{\frac{(x+4)^n}{n(3^n)}} \right| = \left| \frac{(x+4)^{n+1}n(3^n)}{(x+4)^n(n+1)(3^{n+1})} \right| }$

Simplifying the expression, we get:

${ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x+4)n}{3(n+1)} \right| = \frac{|x+4|}{3} \cdot \frac{n}{n+1} }$

Next, we take the limit as n approaches infinity:

${ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{|x+4|}{3} \cdot \frac{n}{n+1} = \frac{|x+4|}{3} \lim_{n \to \infty} \frac{n}{n+1} }$

Since ${\lim_{n \to \infty} \frac{n}{n+1} = 1}$, we have:

${ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{|x+4|}{3} }$

For the series to converge, according to the ratio test, this limit must be less than 1:

${ \frac{|x+4|}{3} < 1 }$

${ |x+4| < 3 }$

This inequality gives us the radius of convergence, ${R = 3}$. To find the interval of convergence, we solve the inequality:

${ -3 < x+4 < 3 }$

${ -7 < x < -1 }$

Now, we need to check the endpoints, ${x = -7}$ and ${x = -1}$, to determine if the series converges at these points.

For ${x = -7}$, the series becomes:

${ \sum_{n=1}^{\infty} \frac{(-7+4)^n}{n(3^n)} = \sum_{n=1}^{\infty} \frac{(-3)^n}{n(3^n)} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} }$

This is an alternating harmonic series, which is known to converge by the alternating series test.

For ${x = -1}$, the series becomes:

${ \sum_{n=1}^{\infty} \frac{(-1+4)^n}{n(3^n)} = \sum_{n=1}^{\infty} \frac{(3)^n}{n(3^n)} = \sum_{n=1}^{\infty} \frac{1}{n} }$

This is the harmonic series, which is known to diverge.

Therefore, the interval of convergence includes ${x = -7}$ but excludes ${x = -1}$. The interval of convergence is ${[-7, -1)}$.

In summary, for the series ${\sum_{n=1}^{\infty} \frac{(x+4)^n}{n(3^n)}}$, the radius of convergence is 3, and the interval of convergence is [-7, -1).

(b) ${\sum_{n=1}^{\infty} \frac{(x-1)^{2n-2}}{(2n-1)!}}$

To find the radius and interval of convergence for the power series ${\sum_{n=1}^{\infty} \frac{(x-1)^{2n-2}}{(2n-1)!}}$, we will again use the ratio test. Let's denote the general term of the series as:

${ a_n = \frac{(x-1)^{2n-2}}{(2n-1)!} }$

Now, we compute the ratio of the (n+1)-th term to the n-th term:

${ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{(x-1)^{2(n+1)-2}}{(2(n+1)-1)!}}{\frac{(x-1)^{2n-2}}{(2n-1)!}} \right| = \left| \frac{(x-1)^{2n}}{(2n+1)!} \cdot \frac{(2n-1)!}{(x-1)^{2n-2}} \right| }$

Simplifying the expression, we get:

${ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x-1)^2}{(2n+1)(2n)} \right| = \frac{|x-1|^2}{(2n+1)(2n)} }$

Next, we take the limit as n approaches infinity:

${ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{|x-1|^2}{(2n+1)(2n)} = |x-1|^2 \lim_{n \to \infty} \frac{1}{(2n+1)(2n)} }$

Since ${\lim_{n \to \infty} \frac{1}{(2n+1)(2n)} = 0}$, we have:

${ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = |x-1|^2 \cdot 0 = 0 }$

For the series to converge, this limit must be less than 1. Since the limit is 0 for all ${x}$, the series converges for all ${x}$. This means the radius of convergence is infinite, ${R = \infty}$, and the interval of convergence is ${(-\infty, \infty)}$.

In summary, for the series ${\sum_{n=1}^{\infty} \frac{(x-1)^{2n-2}}{(2n-1)!}}$, the radius of convergence is infinite, and the interval of convergence is ${(-\infty, \infty)}$.

(c) ${\sum_{n=1}^{\infty} \frac{(-1)^{n-1}(3x-1)^n}{n^2}}$

To determine the radius and interval of convergence for the power series ${\sum_{n=1}^{\infty} \frac{(-1)^{n-1}(3x-1)^n}{n^2}}$, we will once again utilize the ratio test. Let's define the general term of the series as:

${ a_n = \frac{(-1)^{n-1}(3x-1)^n}{n^2} }$

Now, we compute the ratio of the (n+1)-th term to the n-th term:

${ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{(-1)^n(3x-1)^{n+1}}{(n+1)^2}}{\frac{(-1)^{n-1}(3x-1)^n}{n^2}} \right| = \left| \frac{(-1)^n(3x-1)^{n+1}n^2}{(-1)^{n-1}(3x-1)^n(n+1)^2} \right| }$

Simplifying the expression, we get:

${ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)(3x-1)n^2}{(n+1)^2} \right| = |3x-1| \cdot \frac{n^2}{(n+1)^2} }$

Next, we take the limit as n approaches infinity:

${ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} |3x-1| \cdot \frac{n^2}{(n+1)^2} = |3x-1| \lim_{n \to \infty} \frac{n^2}{(n+1)^2} }$

Since ${\lim_{n \to \infty} \frac{n^2}{(n+1)^2} = 1}$, we have:

${ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = |3x-1| }$

For the series to converge, according to the ratio test, this limit must be less than 1:

${ |3x-1| < 1 }$

${ -1 < 3x-1 < 1 }$

${ 0 < 3x < 2 }$

${ 0 < x < \frac{2}{3} }$

To find the radius of convergence, we first rewrite the inequality in the form ${|x - c| < R}$, where ${c}$ is the center and ${R}$ is the radius. We can rewrite ${|3x-1| < 1}$ as:

${ \left| 3\left(x - \frac{1}{3}\right) \right| < 1 }$

${ 3\left| x - \frac{1}{3} \right| < 1 }$

${ \left| x - \frac{1}{3} \right| < \frac{1}{3} }$

Thus, the radius of convergence is ${R = \frac{1}{3}}$. The interval is ${(0, \frac{2}{3})}$. Now, we need to check the endpoints, ${x = 0}$ and ${x = \frac{2}{3}}$, to determine if the series converges at these points.

For ${x = 0}$, the series becomes:

${ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(3(0)-1)^n}{n^2} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(-1)^n}{n^2} = \sum_{n=1}^{\infty} \frac{(-1)^{2n-1}}{n^2} = \sum_{n=1}^{\infty} \frac{-1}{n^2} = -\sum_{n=1}^{\infty} \frac{1}{n^2} }$

This is a constant multiple of the p-series with ${p = 2}$, which is known to converge (p-series test).

For ${x = \frac{2}{3}}$, the series becomes:

${ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(3(\frac{2}{3})-1)^n}{n^2} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(2-1)^n}{n^2} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} }$

This is an alternating series, and the absolute value of the terms, ${\frac{1}{n^2}}$, decreases monotonically to 0. Therefore, by the alternating series test, this series converges.

Thus, the interval of convergence includes both endpoints. The interval of convergence is ${[0, \frac{2}{3}]}$.

In summary, for the series ${\sum_{n=1}^{\infty} \frac{(-1)^{n-1}(3x-1)^n}{n^2}}$, the radius of convergence is ${\frac{1}{3}}$, and the interval of convergence is ${[0, \frac{2}{3}]}$.

(d) ${\sum_{n=1}^{\infty} \frac{x^n}{n^n}}$

To find the radius and interval of convergence for the power series ${\sum_{n=1}^{\infty} \frac{x^n}{n^n}}$, we can use the root test. The root test is particularly effective when dealing with series where the terms involve n-th powers. Let's denote the general term of the series as:

${ a_n = \frac{x^n}{n^n} }$

Now, we compute the n-th root of the absolute value of the general term:

${ \sqrt[n]{|a_n|} = \sqrt[n]{\left| \frac{x^n}{n^n} \right|} = \sqrt[n]{\frac{|x|^n}{n^n}} = \frac{|x|}{n} }$

Next, we take the limit as n approaches infinity:

${ \lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \frac{|x|}{n} = |x| \lim_{n \to \infty} \frac{1}{n} }$

Since ${\lim_{n \to \infty} \frac{1}{n} = 0}$, we have:

${ \lim_{n \to \infty} \sqrt[n]{|a_n|} = |x| \cdot 0 = 0 }$

For the series to converge, according to the root test, this limit must be less than 1. Since the limit is 0 for all ${x}$, the series converges for all ${x}$. This means the radius of convergence is infinite, ${R = \infty}$, and the interval of convergence is ${(-\infty, \infty)}$.

In summary, for the series ${\sum_{n=1}^{\infty} \frac{x^n}{n^n}}$, the radius of convergence is infinite, and the interval of convergence is ${(-\infty, \infty)}$.

In conclusion, determining the radius and interval of convergence is a crucial skill in analyzing power series. The ratio and root tests are powerful tools for this purpose, and careful consideration of the endpoints is essential for establishing the complete interval of convergence. Each series presents its unique challenges, requiring a nuanced approach to ensure accurate results.