4. find the radius of convergence and the interval of convergence for ∑_{k = 1}^∞ (x - 2)^k / k^k

Explanation:

Step1: Apply the ratio - test

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

Step2: Find the radius of convergence

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

Step3: Find the interval of convergence

Solve the inequality $|x - 2|<e$. This gives $-e<x - 2<e$, or $2 - e<x<2 + e$. Check the endpoints: When $x=2 + e$, the series becomes $\sum_{k = 1}^{\infty}\frac{(2 + e-2)^k}{k^k}=\sum_{k = 1}^{\infty}\frac{e^k}{k^k}=\sum_{k = 1}^{\infty}\left(\frac{e}{k}\right)^k$. Since $\lim_{k\rightarrow\infty}\left(\frac{e}{k}\right)^k = 0$ and the series is a positive - term series, and we can use the root - test: $\lim_{k\rightarrow\infty}\sqrt[k]{\left(\frac{e}{k}\right)^k}=\lim_{k\rightarrow\infty}\frac{e}{k}=0<1$, so the series converges at $x = 2+e$. When $x=2 - e$, the series becomes $\sum_{k = 1}^{\infty}\frac{(2 - e-2)^k}{k^k}=\sum_{k = 1}^{\infty}\frac{(-e)^k}{k^k}=\sum_{k = 1}^{\infty}(-1)^k\left(\frac{e}{k}\right)^k$. By the alternating - series test, since $\lim_{k\rightarrow\infty}\left(\frac{e}{k}\right)^k = 0$ and the sequence $b_k=\left(\frac{e}{k}\right)^k$ is decreasing for $k\geq1$, the series converges at $x = 2 - e$.

Answer:

Radius of convergence: $R = e$; Interval of convergence: $[2 - e,2 + e]$