Chapter 9.8, Problem 4QCE

(a) Since $\underset{k\to +\infty }{\lim }\left|\frac{{(x-4)}^{k+1}/\sqrt{k+1}}{{(x-4)}^k/\sqrt{k}}\right|=\underset{k\to +\infty }{\lim }\left|\sqrt{\frac{k}{k+1}}(x-4)\right|=\left|x-4\right|$ the radius of convergence for the infinite series ${\displaystyle {\sum }_{k=1}^{\infty }(1/\sqrt{k}}){(x-4)}^k\text{is \_\_\_\_\_}\text{.}$

(b) When $x=3,$ ${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{1}{\sqrt{k}}}{(x-4)}^k={\displaystyle \underset{k=`1}{\overset{\infty }{\sum }}\frac{1}{\sqrt{k}}}{(-1)}^k$ Does this series converge or diverge?

(c) When $x=5,$ ${\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{1}{\sqrt{k}}{(x-4)}^k={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\frac{1}{\sqrt{k}}}}$ Does this series converge or diverge?

(d) The interval of convergence for the infinite series ${\displaystyle {\sum }_{k=1}^{\infty }(1/\sqrt{k}}){(x-4)}^k\text{is \_\_\_\_\_\_\_}\text{.}$