Science:Math Exam Resources/Courses/MATH101/April 2016/Question 06 (b)/Solution 1

We first apply the Ratio Test with $a_{n}={\frac {(x-1)^{n}}{\sqrt {n}}}$. As

${\begin{aligned}\lim _{n\rightarrow \infty }{\Big |}{\frac {a_{n+1}}{a_{n}}}{\Big |}&=\lim _{n\rightarrow \infty }{\bigg |}{\frac {{(x-1)^{n+1}}/{\sqrt {n+1}}}{{(x-1)^{n}}/{\sqrt {n}}}}{\bigg |}\\&=\lim _{n\rightarrow \infty }{\bigg (}{\frac {\sqrt {n}}{\sqrt {n+1}}}|x-1|{\bigg )}\\&=\lim _{n\rightarrow \infty }{\bigg (}{\frac {1}{\sqrt {1+1/n}}}|x-1|{\bigg )}=|x-1|,\end{aligned}}$

the series converges for $|x-1|<1$ (in other words, for $0<x<2$) and diverges for $|x-1|>1$.

When $x=2$, the series reduces to $\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n}}}$, which is a divergent $p$-series with $p={\frac {1}{2}}$.

On the other hand, when $x=0$, the series reduces to $\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}}$, which converges by the Alternating Series Test.

So the interval of convergence is $[0,2)$.