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

MATH101 April 2018

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Question 08 (i)
Determine the interval of convergence for the power series $$\displaystyle \sum _{n=0}^{\infty }{\frac {(3x-4)^{n}}{2^{n}(2n+1)}}.$$ Explain all the steps in your reasoning.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Use the ratio test.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 ${\textstyle {\color {blue}[2/3,2)}}$.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 ${\textstyle {\color {blue}[2/3,2)}}$.