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

$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{a_{n+1}}{a_n}\right|& =\underset{n\to \mathrm{\infty }}{\lim }\left|\frac{(x-1{)}^{n+1}/\sqrt{n+1}}{(x-1{)}^n/\sqrt{n}}\right|\\ & =\underset{n\to \mathrm{\infty }}{\lim }(\frac{\sqrt{n}}{\sqrt{n+1}}|x-1|)\\ & =\underset{n\to \mathrm{\infty }}{\lim }(\frac{1}{\sqrt{1+1/n}}|x-1|)=|x-1|,\end{array}$

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

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

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