Understanding the Power Series

Oct 13, 2025 by wbrum.org 113 views

Introduction

In mathematics, a power series is a series of the form $\sum_{n=0}^{\infty} a_n (x-c)^n$ , where $a_n$ are coefficients, $x$ is the variable, and $c$ is a constant. The radius of convergence of a power series is the distance from the center of the series to the nearest point where the series diverges. In this article, we will find the radius of convergence of the power series $\sum_{n=1}^{\infty} \frac{(2x-2)^n(5n+3)!}{3n!}$ .

Understanding the Power Series

The given power series is $\sum_{n=1}^{\infty} \frac{(2x-2)^n(5n+3)!}{3n!}$ . To find the radius of convergence, we need to apply the ratio test. The ratio test states that if the limit of the absolute value of the ratio of consecutive terms is less than 1, then the series converges. If the limit is greater than 1, then the series diverges.

Applying the Ratio Test

To apply the ratio test, we need to find the limit of the absolute value of the ratio of consecutive terms. Let's denote the $n^{th}$ term of the series as $a_n$ and the $(n+1)^{th}$ term as $a_{n+1}$ . Then, the ratio of consecutive terms is $\left|\frac{a_{n+1}}{a_n}\right|$ .

\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{\frac{(2x-2)^{n+1}(5(n+1)+3)!}{3(n+1)!}}{\frac{(2x-2)^n(5n+3)!}{3n!}}\right|

Simplifying the expression, we get:

\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(2x-2)^{n+1}(5n+8)!}{3(n+1)!} \cdot \frac{3n!}{(2x-2)^n(5n+3)!}\right|

Cancelling out the common terms, we get:

\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(2x-2)(5n+8)(5n+7)(5n+6)(5n+5)}{3(n+1)}\right|

Finding the Limit

Now, we need to find the limit of the absolute value of the ratio of consecutive terms as $n$ approaches infinity. Let's denote the limit as $L$ .

L = \lim_{n \to \infty} \left|\frac{(2x-2)(5n+8)(5n+7)(5n+6)(5n+5)}{3(n+1)}\right|

To find the limit, we can divide the numerator and denominator by $n^5$ .

L = \lim_{n \to \infty} \left|\frac{(2x-2)\left(1+\frac{8}{n}\right)\left(1+\frac{7}{n}\right)\left(1+\frac{6}{n}\right)\left(1+\frac{5}{n}\right)}{\frac{3}{n^5}(n+1)}\right|

As $n$ approaches infinity, the terms $\frac{8}{n}$ , $\frac{7}{n}$ , $\frac{6}{n}$ , and $\frac{5}{n}$ approach 0.

L = \lim_{n \to \infty} \left|\frac{(2x-2)}{\frac{3}{n^5}(n+1)}\right|

Simplifying the expression, we get:

L = \left|\frac{(2x-2)n^5}{3(n+1)}\right|

Evaluating the Limit

Now, we need to evaluate the limit as $n$ approaches infinity.

L = \lim_{n \to \infty} \left|\frac{(2x-2)n^5}{3(n+1)}\right|

As $n$ approaches infinity, the term $n^5$ grows much faster than the term $n+1$ . Therefore, the limit is:

L = \left|\frac{(2x-2)}{3}\right|

Finding the Radius of Convergence

The radius of convergence is the distance from the center of the series to the nearest point where the series diverges. In this case, the center of the series is $x=1$ . Therefore, the radius of convergence is the distance from $x=1$ to the point where the series diverges.

R = \frac{1}{\left|\frac{2}{3}\right|}

Simplifying the expression, we get:

R = \frac{3}{2}

Conclusion

In this article, we found the radius of convergence of the power series $\sum_{n=1}^{\infty} \frac{(2x-2)^n(5n+3)!}{3n!}$ . The radius of convergence is $\frac{3}{2}$ . This means that the series converges for all values of $x$ within a distance of $\frac{3}{2}$ from the center of the series, which is $x=1$ .

Q: What is the radius of convergence?

A: The radius of convergence is the distance from the center of a power series to the nearest point where the series diverges. It is a measure of how far the series can be extended before it becomes divergent.

Q: How do I find the radius of convergence of a power series?

A: To find the radius of convergence of a power series, you can use the ratio test. The ratio test states that if the limit of the absolute value of the ratio of consecutive terms is less than 1, then the series converges. If the limit is greater than 1, then the series diverges.

Q: What is the ratio test?

A: The ratio test is a method for determining the convergence or divergence of a series. It involves finding the limit of the absolute value of the ratio of consecutive terms. If the limit is less than 1, the series converges. If the limit is greater than 1, the series diverges.

Q: How do I apply the ratio test to a power series?

A: To apply the ratio test to a power series, you need to find the limit of the absolute value of the ratio of consecutive terms. This involves simplifying the expression and evaluating the limit as the term number approaches infinity.

Q: What is the center of a power series?

A: The center of a power series is the value of the variable that makes the series equal to zero. It is the point around which the series is centered.

Q: How do I find the center of a power series?

A: To find the center of a power series, you need to look for the term that makes the series equal to zero. This is usually the term with the highest power of the variable.

Q: What is the significance of the radius of convergence?

A: The radius of convergence is significant because it tells us how far a power series can be extended before it becomes divergent. It is a measure of the stability of the series.

Q: Can a power series have a radius of convergence of zero?

A: Yes, a power series can have a radius of convergence of zero. This means that the series diverges for all values of the variable.

Q: Can a power series have a radius of convergence of infinity?

A: No, a power series cannot have a radius of convergence of infinity. This is because the series would converge for all values of the variable, which is not possible.

Q: How do I use the radius of convergence to determine the convergence of a power series?

A: To use the radius of convergence to determine the convergence of a power series, you need to compare the distance of the variable from the center of the series to the radius of convergence. If the distance is less than the radius of convergence, the series converges. If the distance is greater than the radius of convergence, the series diverges.

Q: Can the radius of convergence be negative?

A: No, the radius of convergence cannot be negative. This is because the radius of convergence is a measure of distance, and distance cannot be negative.

Q: Can the radius of convergence be complex?

A: No, the radius of convergence cannot be complex. This is because the radius of convergence is a measure of distance, and distance is always real.

Q: How do I find the radius of convergence of a power series with complex coefficients?

A: To find the radius of convergence of a power series with complex coefficients, you need to use the same method as for power series with real coefficients. However, you will need to use complex analysis techniques to evaluate the limit.

Q: Can the radius of convergence be a function of the variable?

A: No, the radius of convergence cannot be a function of the variable. This is because the radius of convergence is a constant that depends only on the coefficients of the power series.

Q: Can the radius of convergence be a function of the coefficients?

A: Yes, the radius of convergence can be a function of the coefficients. This is because the radius of convergence depends on the coefficients of the power series.

Q: How do I use the radius of convergence to determine the convergence of a power series with variable coefficients?

A: To use the radius of convergence to determine the convergence of a power series with variable coefficients, you need to compare the distance of the variable from the center of the series to the radius of convergence. If the distance is less than the radius of convergence, the series converges. If the distance is greater than the radius of convergence, the series diverges.

Q: Can the radius of convergence be used to determine the convergence of a power series with complex coefficients?

A: Yes, the radius of convergence can be used to determine the convergence of a power series with complex coefficients. However, you will need to use complex analysis techniques to evaluate the limit.

Q: How do I use the radius of convergence to determine the convergence of a power series with complex coefficients and variable coefficients?

A: To use the radius of convergence to determine the convergence of a power series with complex coefficients and variable coefficients, you need to compare the distance of the variable from the center of the series to the radius of convergence. If the distance is less than the radius of convergence, the series converges. If the distance is greater than the radius of convergence, the series diverges.