Science:Math Exam Resources/Courses/MATH101/April 2017/Question 06 (b)/Solution 1

Apply the ratio test. Since we have

$$a_{n}={\frac {2}{3^{n}(x-1)^{n+1}}}.$$

and

$${\frac {a_{n+1}}{a_{n}}}={\frac {2}{3^{n+1}(x-1)^{n+2}}}\cdot {\frac {3^{n}(x-1)^{n+1}}{2}}={\frac {1}{3(x-1)}},$$

the following three situations are given

$$\lim _{n\rightarrow \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|{\begin{cases}>1,&{\text{the series diverges}}\\=1,&{\text{inconclusive}}\\<1,&{\text{the series converges}}\\\end{cases}}$$

Therefore, for ${\textstyle |x-1|>{\frac {1}{3}}}$, we have ${\textstyle \lim _{n\rightarrow \infty }|{\frac {a_{n+1}}{a_{n}}}|={\frac {1}{3|x-1|}}<1}$, so that the series converges.

Now, we determine the convergence of series when $|x-1|={\frac {1}{3}}$.

When $x-1={\frac {1}{3}}$ and $x-1=-{\frac {1}{3}}$ , we have

$a_{n}={\frac {2}{3^{n}(x-1)^{n+1}}}={\frac {2}{3^{n}\left({\frac {1}{3}}\right)^{n+1}}}=6$

and

$a_{n}={\frac {2}{3^{n}(x-1)^{n+1}}}={\frac {2}{3^{n}\left(-{\frac {1}{3}}\right)^{n+1}}}=(-1)^{n+1}6$, respectively.

Since in both case, $\lim _{n\to \infty }a_{n}\neq 0$, by the divergence test, the series doesn't converges.

To summarize, the series converges on ${\textstyle \color {blue}\{x:|x-1|>{\frac {1}{3}}\}=\{x<{\frac {2}{3}}{\text{ or }}x>{\frac {4}{3}}\}.}$