Question

Find the radius of convergence of the power series. (if you need to use or –, enter infinity or –infinity, respectively. ) [infinity] (−1)n xn 2n n = 0

Answer 1

For given power series $\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^n}$ the radius of convergence is 2.

For given question,

We have been given a power series $\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^n}$

We need to find the radius of convergence of the power series.

We use ratio test to find the radius of convergence of the power series.

Let $a_n=\frac{(-1)^nx^n}{2^n}$

$\Rightarrow a_{n+1}=\frac{(-1)^{n+1}x^{n+1}}{2^{n+1}}$

Consider,

$\lim_{n \to \infty}|\frac{a_{n+1}}{a_n} |\\\\= \lim_{n \to \infty} |\frac{\frac{(-1)^{n+1}x^{n+1}}{2^{n+1}}}{ \frac{(-1)^nx^n}{2^n} } |\\\\=\lim_{n \to \infty} |\frac{(-1)^{n+1}x^{n+1}}{2^{n+1}}\times \frac{2^n}{(-1)^nx^n} |\\\\=\lim_{n \to \infty} |\frac{(-1)x}{2} |\\\\=\lim_{n \to \infty}|\frac{-x}{2} |\\\\=\frac{x}{2}$

By Ratio test, given power series converges at $|\frac{x}{2} | < 1$

⇒ |x| < 2

So, the radius of convergence is 2.

Therefore, for given power series $\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{2^n}$ the radius of convergence is 2.

Learn more about the radius of convergence here:

brainly.com/question/2289050

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