Science:Math Exam Resources/Courses/MATH101/April 2018/Question 08 (i)/Solution 1

Using ${\textstyle a_{n}={\frac {(3x-4)^{n}}{2^{n}(2n+1)}}}$, the relevant ratio for the ratio test is

$$\left|{\frac {a_{n+1}}{a_{n}}}\right|=\left|{\frac {(3x-4)^{n+1}}{2^{n+1}(2n+3)}}\cdot {\frac {2^{n}(2n+1)}{(3x-4)^{n}}}\right|=\left|{\frac {3x-4}{2}}\cdot {\frac {2n+1}{2n+3}}\right|.$$

Observe that

$$\lim _{n\to \infty }{\frac {2n+1}{2n+3}}=\lim _{n\to \infty }{\frac {2+{\frac {1}{n}}}{2+{\frac {3}{n}}}}=1,$$

so we have ${\textstyle \lim _{n\to \infty }|a_{n+1}/a_{n}|=\left|{\frac {3x-4}{2}}\right|.}$ Now the ratio test says that the series converges when ${\textstyle |(3x-4)/2|<1}$ or equivalently (after solving for ${\textstyle x}$), when ${\textstyle 2/3<x<2}$; while the series diverges when ${\textstyle |(3x-4)/2|>1}$. i.e., $x<2/3$ or $x>2$. It remains to check the endpoints.

When ${\textstyle x=2}$, the summand is

$${\frac {2^{n}}{2^{n}(2n+1)}}={\frac {1}{2n+1}}\geq {\frac {1}{3n}}$$

for ${\textstyle n\geq 1}$, which diverges by comparison test with the harmonic series. On the other hand, when ${\textstyle x=2/3}$ the summand is

$${\frac {(-2)^{n}}{2^{n}(2n+1)}}=(-1)^{n}{\frac {1}{2n+1}}$$

which converges by the alternating series test.

Answer: The correct answer is ${\textstyle {\color {blue}[2/3,2)}}$.