attempt 1: 10 attempts remaining. find the radius and interval of convergence of the series sum from k = 1…

Explanation:

Step1: Apply the ratio - test

Let $a_k=\frac{k^2x^{2k}}{2^{2k}}$. Then $a_{k + 1}=\frac{(k + 1)^2x^{2(k+1)}}{2^{2(k + 1)}}=\frac{(k + 1)^2x^{2k+2}}{2^{2k+2}}$. Calculate the limit $L=\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_k}\right|$. [ \begin{align*} L&=\lim_{k\rightarrow\infty}\left|\frac{\frac{(k + 1)^2x^{2k+2}}{2^{2k+2}}}{\frac{k^2x^{2k}}{2^{2k}}}\right|\ &=\lim_{k\rightarrow\infty}\left|\frac{(k + 1)^2x^{2k+2}}{2^{2k+2}}\cdot\frac{2^{2k}}{k^2x^{2k}}\right|\ &=\lim_{k\rightarrow\infty}\left|\frac{(k + 1)^2}{k^2}\cdot\frac{x^{2k+2}}{x^{2k}}\cdot\frac{2^{2k}}{2^{2k+2}}\right|\ &=\lim_{k\rightarrow\infty}\left(\left(\frac{k + 1}{k}\right)^2\cdot|x|^2\cdot\frac{1}{4}\right) \end{align*} ] Since $\lim_{k\rightarrow\infty}\frac{k + 1}{k}=1$, then $L=\frac{|x|^2}{4}$.

Step2: Find the radius of convergence

For convergence, $L<1$. So $\frac{|x|^2}{4}<1$, which implies $|x|^2 < 4$, or $|x|<2$. The radius of convergence $R = 2$.

Step3: Check the endpoints

When $x = 2$, the series becomes $\sum_{k = 1}^{\infty}\frac{k^2\cdot2^{2k}}{2^{2k}}=\sum_{k = 1}^{\infty}k^2$. By the $n$ - th term test for divergence, $\lim_{k\rightarrow\infty}k^2=\infty$, so the series diverges at $x = 2$. When $x=-2$, the series becomes $\sum_{k = 1}^{\infty}\frac{k^2(-2)^{2k}}{2^{2k}}=\sum_{k = 1}^{\infty}k^2$, which also diverges.

Answer:

$R = 2$ Interval of convergence $=(-2,2)$