determine the radius and interval of convergence of the following power series.\n sum_{k =…

Explanation:

Step1: Apply the ratio - test

Let $a_k=\frac{k^{2}(x + 2)^{2k}}{k!}$. Then $a_{k+1}=\frac{(k + 1)^{2}(x + 2)^{2(k+1)}}{(k+1)!}$. Calculate the limit $L=\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_k}\right|$. [ \begin{align*} \frac{a_{k+1}}{a_k}&=\frac{\frac{(k + 1)^{2}(x + 2)^{2(k + 1)}}{(k+1)!}}{\frac{k^{2}(x + 2)^{2k}}{k!}}\ &=\frac{(k + 1)^{2}(x + 2)^{2k+2}k!}{(k+1)!k^{2}(x + 2)^{2k}}\ &=\frac{(k + 1)^{2}(x + 2)^{2}}{(k + 1)k^{2}}\ &=\frac{(k + 1)(x + 2)^{2}}{k^{2}} \end{align*} ]

Step2: Find the limit

[ \begin{align*} L&=\lim_{k\rightarrow\infty}\left|\frac{(k + 1)(x + 2)^{2}}{k^{2}}\right|\ &=(x + 2)^{2}\lim_{k\rightarrow\infty}\frac{k+1}{k^{2}}\ &=(x + 2)^{2}\lim_{k\rightarrow\infty}\left(\frac{1}{k}+\frac{1}{k^{2}}\right)\ &=0 \end{align*} ] Since $L = 0$ for all real - valued $x$, the radius of convergence $R=\infty$.

Step3: Determine the interval of convergence

Since the radius of convergence $R=\infty$, the interval of convergence is $(-\infty,\infty)$.

Answer:

Radius of convergence $R = \infty$, Interval of convergence $(-\infty,\infty)$