consider the power series\n∑(k = 1 to ∞) k^k(x - 5)^k.\nfind the radius of convergence r.\nr =\nwhat is the…

Explanation:

Step1: Apply ratio - test formula

Let $a_k=k^k(x - 5)^k$. Then $\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_k}\right|=\lim_{k\rightarrow\infty}\left|\frac{(k + 1)^{k+1}(x - 5)^{k+1}}{k^k(x - 5)^k}\right|$.

Step2: Simplify the ratio

$\lim_{k\rightarrow\infty}\left|\frac{(k + 1)^{k+1}(x - 5)^{k+1}}{k^k(x - 5)^k}\right|=\lim_{k\rightarrow\infty}\left|\left(\frac{k + 1}{k}\right)^k(k + 1)(x - 5)\right|$. Since $\lim_{k\rightarrow\infty}\left(\frac{k+1}{k}\right)^k = e$, the limit becomes $\lim_{k\rightarrow\infty}|e(k + 1)(x - 5)|$.

Step3: Find the radius of convergence

For convergence, $\lim_{k\rightarrow\infty}|e(k + 1)(x - 5)|<1$. But as $k\rightarrow\infty$, this is only true when $x = 5$. We can also use the root - test. $\lim_{k\rightarrow\infty}\sqrt[k]{|a_k|}=\lim_{k\rightarrow\infty}\sqrt[k]{|k^k(x - 5)^k|}=\lim_{k\rightarrow\infty}|k(x - 5)|$. For convergence, $\lim_{k\rightarrow\infty}|k(x - 5)|<1$. This is true when $x = 5$. The radius of convergence $R$ is found by considering $\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_k}\right|$. We know that for a power series $\sum a_k(x - c)^k$, the radius of convergence $R$ is given by $\frac{1}{L}$ where $L=\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_k}\right|$ (when this limit exists). Here, $\lim_{k\rightarrow\infty}\left|\frac{(k + 1)^{k+1}}{k^k}\right|=\lim_{k\rightarrow\infty}(k + 1)\left(\frac{k + 1}{k}\right)^k=\infty$. So, $R = 0$.

Step4: Find the interval of convergence

Since $R = 0$, the power series converges only at the center $x=5$. The interval of convergence is ${5}$.

Answer:

$R = 0$ Interval of convergence: ${5}$