Science:Math Exam Resources/Courses/MATH101/April 2018/Question 08 (i)/Solution 2

In this alternative solution (outside the curriculum), we will use the root test.

The first step is to manipulate the summand so that the series actually looks like a power series. If we divide both the numerator and the denominator by ${\textstyle 3^{n}}$, then we get

$${\frac {(3x-4)^{n}/3^{n}}{(2/3)^{n}(2n+1)}}={\frac {{\big (}(3x-4)/3{\big )}^{n}}{(2/3)^{n}(2n+1)}}={\frac {{\big (}x-{\frac {4}{3}}{\big )}^{n}}{(2/3)^{n}(2n+1)}}.$$

Now we see that the series has the form

$$\sum _{n=1}^{\infty }c_{n}(x-a)^{n}$$

with ${\textstyle a=4/3}$ and

$$c_{n}={\frac {1}{(2/3)^{n}(2n+1)}}.$$

By definition, the reciprocal of the radius of convergence ${\textstyle R}$ of the series is given by

$$1/R=\limsup _{n\to \infty }|c_{n}|^{1/n}=\limsup _{n\to \infty }{\frac {1}{{\frac {2}{3}}\cdot (2n+1)^{1/n}}}.$$

Notice that for any positive integer ${\textstyle n}$ we have

$$(2n+1)^{1/n}\geq (2n)^{1/n}=2^{1/n}\cdot n^{1/n}$$

and

$$(2n+1)^{1/n}\leq (3n)^{1/n}=3^{1/n}\cdot n^{1/n}.$$

It follows that

$$\limsup _{n\to \infty }{\frac {1}{{\frac {2}{3}}\cdot (2n+1)^{1/n}}}\leq \limsup _{n\to \infty }{\frac {1}{{\frac {2}{3}}\cdot 2^{1/n}\cdot n^{1/n}}}={\frac {1}{{\frac {2}{3}}\cdot 1}}=3/2$$

and

$$\limsup _{n\to \infty }{\frac {1}{{\frac {2}{3}}\cdot (2n+1)^{1/n}}}\geq \limsup _{n\to \infty }{\frac {1}{{\frac {2}{3}}\cdot 3^{1/n}\cdot n^{1/n}}}={\frac {1}{{\frac {2}{3}}\cdot 1}}=3/2$$

so ${\textstyle 1/R=3/2}$ by the squeeze theorem. This shows that the radius of convergence of the power series (centred at ${\textstyle a=4/3}$) is equal to ${\textstyle 2/3}$, which implies that the series convergence for ${\textstyle 2/3<x<2}$. It remains to check the endpoints of this interval.

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 ${\textstyle {\color {blue}[2/3,2)}}$.