Answer
$a) R= 1$ ,interval of convergence : $0\leq x \lt2$
$b)$ interval of absolute convergence : $0\lt x \lt2$
$c)$ series converges conditionally $x = 0$
Work Step by Step
Step 1.
Use the Ratio Test to find the interval where the series converges absolutely.
$\frac{{u_{n+1} }}{{u_{n} }} = \frac{(x-1)^{n+1} }{\sqrt{n+1}} \times \frac{\sqrt n}{(x-1)^{n}} = |\frac{(x-1)\times\sqrt n}{\sqrt {n+1}}| = \lim\limits_{n \to \infty} |x-1| \times \lim\limits_{n \to \infty} |\frac{\sqrt n}{\sqrt {n+1}}| = |x-1|
$
$ -1 < x-1 < 1 \\
0 < x < 2 \\$
the radius is $R=(2-0) : 2 = 1 \\ R = 1$
the interval of absolute convergence is $0 < x<2 $
Step 2.
Check points
$1) x = 0 ⇒ \Sigma \frac{(0-1)^n}{\sqrt n} = \frac{(-1)^n}{\sqrt n} ⇒ $ series is conditionally convergent ⇒ $x = 0$ $include$.
$2) x = 2 ⇒ \Sigma \frac{(2-1)^n}{\sqrt n} = \frac{(1)^n}{\sqrt n} = \frac{1}{\sqrt n} ⇒ $ series is divergent ⇒ $x = 2$ $not $ $include$.
Then, interval of convergence is $0\leq x \lt2$
So, answer
$a) R= 1,interval$ $of$ $convergence : $ $0\leq x \lt2$
$b)$ $interval$ $of$ $absolute$ $convergence : $ $0\lt x \lt2$
$c)$ $series$ $converges$ $conditionally $ $x = 0$
