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].
