determine the radius and interval of convergence of the following power series. \n∑_{k = 1}^∞ (x^{2k +…

Explanation:

Step1: Apply the ratio - test

Let $a_k=\frac{x^{2k + 1}}{2^{k-1}}$. Then $a_{k + 1}=\frac{x^{2(k + 1)+1}}{2^{(k + 1)-1}}=\frac{x^{2k+3}}{2^{k}}$. Calculate $\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|=\lim_{k\rightarrow\infty}\left|\frac{\frac{x^{2k + 3}}{2^{k}}}{\frac{x^{2k+1}}{2^{k - 1}}}\right|$.

Step2: Simplify the ratio

$\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|=\lim_{k\rightarrow\infty}\left|\frac{x^{2k+3}}{2^{k}}\cdot\frac{2^{k - 1}}{x^{2k+1}}\right|=\lim_{k\rightarrow\infty}\left|\frac{x^{2k + 3-2k - 1}}{2^{k-(k - 1)}}\right|=\lim_{k\rightarrow\infty}\left|\frac{x^{2}}{2}\right|=\frac{|x|^{2}}{2}$.

Step3: Find the radius of convergence

For convergence, $\frac{|x|^{2}}{2}<1$. Solving $|x|^{2}<2$, we get $|x|<\sqrt{2}$. The radius of convergence $R$ is given by the formula for the ratio - test result set to less than 1. Since $\frac{|x|^{2}}{2}<1$ implies $|x|<\sqrt{2}$, the radius of convergence $R = \sqrt{2}$.

Answer:

$\sqrt{2}$