Determining the Interval of Convergence

Oct 13, 2025 by thesweetdesigns.com 113 views

Determine the Interval of Convergence for the Series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$

The interval of convergence is a crucial concept in the study of power series, which is a series of the form $\sum_{n=0}^{\infty} a_n (x-c)^n$ . In this article, we will focus on determining the interval of convergence for the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ .

Understanding the Series

The given series is a power series with the general term $\frac{(-1)^n x^{2n}}{n!}$ . The series has a few notable features:

The series has a negative sign in front of the term, which is raised to the power of $n$ . This means that the series will alternate between positive and negative terms.
The series has a term $x^{2n}$ , which means that the power of $x$ increases by 2 with each term.
The series has a factorial term $n!$ in the denominator, which means that the denominator will grow rapidly as $n$ increases.

Determining the Interval of Convergence

To determine the interval of convergence, we need to find the values of $x$ for which the series converges. One way to do this is to use the ratio test, which states that a series $\sum_{n=1}^{\infty} a_n$ converges if the limit of $\left|\frac{a_{n+1}}{a_n}\right|$ as $n$ approaches infinity is less than 1.

Let's apply the ratio test to the given series:

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

Simplifying the expression, we get:

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

Taking the limit as $n$ approaches infinity, we get:

\lim_{n\to\infty} \left|\frac{(-1) x^{2}}{(n+1)}\right| = 0

Since the limit is 0, which is less than 1, the series converges for all values of $x$ .

Finding the Radius of Convergence

The radius of convergence is the distance from the center of the series to the point where the series diverges. In this case, the series converges for all values of $x$ , so the radius of convergence is infinite.

Finding the Interval of Convergence

Since the series converges for all values of $x$ , the interval of convergence is the entire real line, which can be written as $(-\infty, \infty)$ .

Conclusion

In this article, we determined the interval of convergence for the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ . We used the ratio test to show that the series converges for all values of $x$ , and therefore the interval of convergence is the entire real line, $(-\infty, \infty)$ .

Answer

The correct answer is C. $(-\infty,+\infty)$ .

References

[1] "Power Series" by Wolfram MathWorld
[2] "Ratio Test" by Wolfram MathWorld
[3] "Interval of Convergence" by Wolfram MathWorld
Determine the Interval of Convergence for the Series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ : Q&A

In our previous article, we determined the interval of convergence for the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ . In this article, we will answer some common questions related to the series and its interval of convergence.

Q: What is the interval of convergence for the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ ?

A: The interval of convergence for the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ is the entire real line, $(-\infty, \infty)$ .

Q: Why does the series converge for all values of $x$ ?

A: The series converges for all values of $x$ because the limit of the ratio test is 0, which is less than 1. This means that the series is absolutely convergent for all values of $x$ .

Q: What is the radius of convergence for the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ ?

A: The radius of convergence for the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ is infinite, since the series converges for all values of $x$ .

Q: Can the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ be used to approximate a function?

A: Yes, the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ can be used to approximate a function. In fact, the series is a power series, which is a series of the form $\sum_{n=0}^{\infty} a_n (x-c)^n$ . Power series are often used to approximate functions, and the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ is a specific example of a power series.

Q: How can I use the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ to approximate a function?

A: To use the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ to approximate a function, you can use the following steps:

Choose a value of $x$ for which you want to approximate the function.
Plug the value of $x$ into the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ .
Calculate the sum of the series up to a certain number of terms.
Use the sum of the series as an approximation of the function.

Q: What are some common functions that can be approximated using the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ ?

A: Some common functions that can be approximated using the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ include:

The exponential function $e^x$
The sine function $\sin x$
The cosine function $\cos x$

Conclusion

In this article, we answered some common questions related to the series $\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n}}{n!}$ and its interval of convergence. We also discussed how to use the series to approximate a function, and provided some examples of common functions that can be approximated using the series.

References

[1] "Power Series" by Wolfram MathWorld
[2] "Ratio Test" by Wolfram MathWorld
[3] "Interval of Convergence" by Wolfram MathWorld
[4] "Approximating Functions with Power Series" by Wolfram MathWorld