91Ó°ÊÓ

#tag_title#Step 2: Apply the Ratio Test#a #tag_content#The Ratio Test states that for any power series, we can find the limit as k approaches infinity of the absolute value of the ratio between two consecutive terms as follows: $$\lim _{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|$$ For our power series, compute the ratio of consecutive terms: $$\frac{a_{k+1}}{a_k}=\frac{(-1)^{k+1}\frac{x^{3(k+1)}}{27^{k+1}}}{(-1)^k\frac{x^{3k}}{27^k}}$$ Simplify the expression: $$\frac{a_{k+1}}{a_k}=-\frac{x^{3k+3}}{27x^{3k}}$$ Now, find the limit as k approaches infinity: $$L=\lim _{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty}\left|\frac{x^{3k+3}}{27x^{3k}}\right|=\left|\frac{x^3}{27}\right|$$ #tag_title#Step 3: Determine the radius of convergence#a #tag_content#For the series to converge, the Ratio Test requires the limit to be less than 1: $$\left|\frac{x^3}{27}\right|<1$$ To find the radius of convergence, solve for x: $$|x^3|<27$$ $$|x|<3$$ Thus, the radius of convergence (R) is 3. #tag_title#Step 4: Determine the interval of convergence (IOC)#a #tag_content#To find the interval of convergence, we need to check the convergence of the power series at the endpoints of the interval |x| < 3, which are x = -3 and x = 3. For x = -3, we have the series: $$\sum_{k=0}^{\infty}(-1)^{k}\frac{(-3)^{3k}}{27^k}=\sum_{k=0}^{\infty}(-1)^k$$ This is an alternating series that does not converge. For x = 3, we have: $$\sum_{k=0}^{\infty}(-1)^{k}\frac{(3)^{3k}}{27^k}=\sum_{k=0}^{\infty} (-1)^k$$ This is also an alternating series that does not converge. Since the series does not converge at the endpoints, the interval of convergence is: $$-3<x<3$$ #Answer# The radius of convergence (R) is 3 and the interval of convergence is (-3, 3).