Understanding the Radius and Interval of Convergence of a Power Series

In this article, we will explore the process of finding the radius of convergence and the interval of convergence for a given power series. Specifically, we will focus on the series (sum_{n1}^{infty} frac{e^{-3n} n!}{n^n} (1 - x^{2n})).

Introduction to Convergence Tests

One of the most effective ways to determine the convergence of a series, especially a power series, is by using various convergence tests. Among these, the Cauchy’s root test and the Ratio Test are particularly useful. This article will explain how to apply these tests to find the radius of convergence and the interval of convergence for the given power series.

Applying Cauchy’s Root Test

Let's apply Cauchy’s root test to find the radius of convergence (R) of the given series:

(sum_{n1}^{infty} frac{e^{-3n} n!}{n^n} (1 - x^{2n}) sum_{n1}^{infty} frac{e^{-3n} n!}{n^n} x^n sum a_n x^n)

Now, we follow the steps of Cauchy’s root test:

First, calculate the limit:

lim_{n to infty} (a_n)^{1/n}

Substitute (a_n frac{e^{-3n} n!}{n^n}) into the expression:

lim_{n to infty} left( frac{e^{-3n} n!}{n^n} right)^{1/n}  e^{-3} lim_{n to infty} left( frac{n!}{n^n} right)^{1/n}

Apply Stirling's approximation to simplify the expression:

 e^{-3} lim_{n to infty} left( frac{n^{n-1/2} e^{-n} sqrt{2 pi n}}{n^n} right)^{1/n}  e^{-3} lim_{n to infty} left( frac{e^{-n} sqrt{2 pi n}}{n^{n-1/2}} right)

Further simplify the expression:

 e^{-3} lim_{n to infty} left( frac{sqrt{2 pi n}}{e^n n^{n-1/2}} right)  e^{-3} lim_{n to infty} left( frac{sqrt{2 pi}}{e n^{1-1/2}} right)  e^{-3} lim_{n to infty} left( frac{sqrt{2 pi}}{e n^{1/2}} right)

Finally, we get:

(e^{-3} lim_{n to infty} left( frac{sqrt{2 pi} n^{1/2}}{e n} right) e^{-3} cdot frac{sqrt{2 pi}}{e} e^{-4} cdot sqrt{2 pi})

In conclusion, the radius of convergence (R) is:

(frac{1}{e^2})

Using the Ratio Test for a More Direct Approach

To verify and refine the radius of convergence, we can use the Ratio Test. The Ratio Test involves calculating the limit of the ratio of consecutive terms:

( r lim_{n to infty} left| frac{a_{n 1}}{a_n} right| )

For the given series:

(a_n frac{e^{-3n} n!}{n^n} x^n)

Calculate the ratio:

(
r  lim_{n to infty} left| frac{e^{-3(n 1)} (n 1)! (1 - x^{2(n 1)}) (1 - x^2)^n}{(1 - x^{2n}) (n 1)! e^{-3n} n^n} right|  lim_{n to infty} |e^{-3} x - 1|^2 cdot lim_{n to infty} frac{(n 1)^{n 1}}{n^n}
)

Simplify the expression:

 |e^{-3} x - 1|^2 cdot lim_{n to infty} left( 1   frac{1}{n} right)^n  |e^{-3} x - 1|^2 cdot e
)

The series converges if:

(|e^{-3} x - 1|^2 e 1) or equivalently (|x - e^2| e^2)

Hence, the radius of convergence is:

(R e^2)

Interval of Convergence

To determine the interval of convergence, we need to check the convergence of the series at the endpoints (x pm e^2).

Testing the Endpoints

1. **For (x e^2):

The series reduces to:

(sum_{n1}^{infty} frac{e^n n!}{n^n})

Using Stirling's approximation:

(lim_{n to infty} frac{e^n n!}{n^n}  lim_{n to infty} frac{e^n n^{n-1/2} e^{-n} sqrt{2 pi n}}{n^n}  lim_{n to infty} sqrt{2 pi n}  infty)

Since the limit is non-zero, the series diverges at (x e^2).

2. **For (x -e^2):

The series is similar but with the alternating sign, which also results in divergence for the same reason.

Therefore, the interval of convergence is:

((-e^2, e^2))

Conclusion

In this article, we discussed the process of finding the radius and interval of convergence for a given power series using Cauchy’s root test and the Ratio Test. We determined the radius of convergence to be (e^2) and concluded that the series converges for (-e^2 x e^2).