Answer :

To find the radius of convergence, [tex]\(R\)[/tex], of the series

[tex]\[ \sum_{n=1}^{\infty} 4(-1)^n n x^n, \][/tex]

we can use the Ratio Test. The Ratio Test is particularly useful for determining the radius of convergence of a power series.

Let's denote the [tex]\(n\)[/tex]-th term of the series by [tex]\(a_n\)[/tex]. Then

[tex]\[ a_n = 4(-1)^n n x^n. \][/tex]

According to the Ratio Test, we need to compute the limit

[tex]\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \][/tex]

First, we express [tex]\(a_{n+1}\)[/tex]:

[tex]\[ a_{n+1} = 4(-1)^{n+1} (n+1) x^{n+1}. \][/tex]

Next, we form the ratio [tex]\(\left| \frac{a_{n+1}}{a_n} \right|\)[/tex]:

[tex]\[ \frac{a_{n+1}}{a_n} = \frac{4(-1)^{n+1} (n+1) x^{n+1}}{4(-1)^n n x^n}. \][/tex]

Simplifying this, we get:

[tex]\[ \frac{a_{n+1}}{a_n} = \frac{4(-1)^{n+1} (n+1) x^{n+1}}{4(-1)^n n x^n} = \frac{(-1)^{n+1} (n+1) x^{n+1}}{(-1)^n n x^n}. \][/tex]

Since [tex]\((-1)^{n+1} / (-1)^n = -1\)[/tex] and [tex]\(x^{n+1} / x^n = x\)[/tex], this further simplifies to:

[tex]\[ \frac{a_{n+1}}{a_n} = \frac{-1 (n+1) x}{n} = - \frac{(n+1) x}{n}. \][/tex]

Thus,

[tex]\[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(n+1) x}{n} \right| = \left| x \right| \frac{n+1}{n} = \left| x \right| \left( 1 + \frac{1}{n} \right). \][/tex]

We now take the limit as [tex]\(n\)[/tex] approaches infinity:

[tex]\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| x \right| \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right) = \left| x \right| \cdot 1 = \left| x \right|. \][/tex]

According to the Ratio Test, the series converges when this limit is less than 1:

[tex]\[ \left| x \right| < 1. \][/tex]

Hence, the radius of convergence [tex]\(R\)[/tex] is:

[tex]\[ R = 1. \][/tex]

The radius of convergence of the series [tex]\(\sum_{n=1}^{\infty} 4(-1)^n n x^n\)[/tex] is [tex]\(R = 1\)[/tex].