consider the power series $sum_{n = 1}^{infty}n^{n}(x - 8)^{n}$. find the radius of convergence $r$. if it…

Explanation:

Step1: Apply ratio - test formula

Let $a_n=n^n(x - 8)^n$. Then $a_{n+1}=(n + 1)^{n+1}(x - 8)^{n+1}$. The ratio - test formula is $\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_n}\right|=\lim_{n\rightarrow\infty}\left|\frac{(n + 1)^{n+1}(x - 8)^{n+1}}{n^n(x - 8)^n}\right|$.

Step2: Simplify the ratio

$\lim_{n\rightarrow\infty}\left|\frac{(n + 1)^{n+1}(x - 8)^{n+1}}{n^n(x - 8)^n}\right|=\lim_{n\rightarrow\infty}\left|\left(\frac{n + 1}{n}\right)^n(n + 1)(x - 8)\right|$. Since $\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^n = e$, the limit becomes $\lim_{n\rightarrow\infty}|e(n + 1)(x - 8)|$.

Step3: Find the radius of convergence

For convergence, $\lim_{n\rightarrow\infty}|e(n + 1)(x - 8)|<1$. But as $n\rightarrow\infty$, the only way for this limit to be less than 1 is when $x - 8 = 0$, i.e., $x=8$. Using the formula $R=\frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}}$, where $a_n=n^n$, $\limsup_{n\rightarrow\infty}\sqrt[n]{|n^n|}=\lim_{n\rightarrow\infty}n=\infty$. So $R = 0$.

Step4: Determine the interval of convergence

Since the radius of convergence $R = 0$, the power - series converges only at the center point. The center of the power series $\sum_{n = 1}^{\infty}n^n(x - 8)^n$ is $x = 8$. So the interval of convergence is $[8,8]$.

Answer:

The radius of convergence $R = 0$ and the interval of convergence is $[8,8]$. Both answers are correct.