attempt 1: 10 attempts remaining. consider the power series (sum_{k = 1}^{infty}\frac{2^{k}(x…

Explanation:

Step1: Apply ratio - test

Let $a_k=\frac{2^{k}(x - 13)^{2k+1}}{(k + 1)!}$. Then $a_{k+1}=\frac{2^{k+1}(x - 13)^{2(k + 1)+1}}{(k+2)!}=\frac{2^{k+1}(x - 13)^{2k + 3}}{(k + 2)!}$. Calculate the limit $L=\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_k}\right|$. [ \begin{align*} \left|\frac{a_{k+1}}{a_k}\right|&=\left|\frac{\frac{2^{k+1}(x - 13)^{2k+3}}{(k + 2)!}}{\frac{2^{k}(x - 13)^{2k+1}}{(k + 1)!}}\right|\ &=\left|\frac{2^{k+1}(x - 13)^{2k+3}(k + 1)!}{2^{k}(x - 13)^{2k+1}(k + 2)!}\right|\ &=\left|\frac{2(x - 13)^{2}}{k+2}\right| \end{align*} ]

Step2: Find the limit as $k\rightarrow\infty$

$\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_k}\right|=\lim_{k\rightarrow\infty}\left|\frac{2(x - 13)^{2}}{k+2}\right| = 0$ for all $x\in\mathbb{R}$.

Answer:

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