Science:Math Exam Resources/Courses/MATH105/April 2015/Question 06 (b)

MATH105 April 2015

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Question 06 (b)
Suppose that the series $\sum \limits _{n=0}^{\infty }(1-a_{n})$ converges, where $a_{n}>0$ for $n=0,1,2,3,\dots .$ Find the radius of convergence of the power series $\sum \limits _{n=0}^{\infty }a_{n}x^{n}$.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1

The radius of convergence of a power series (centred at 0) is the number $R$ for which the power series converges for all $|x|<R$ but diverges for all $|x|>R$. The ratio test can be applied to determine the radius of convergence of such a power series: it says that the series converges when $\lim _{n\rightarrow \infty }\left|{\frac {a_{n+1}x^{n+1}}{a_{n}x^{n}}}\right|<1$, and diverges when $\lim _{n\rightarrow \infty }\left|{\frac {a_{n+1}x^{n+1}}{a_{n}x^{n}}}\right|>1$.

Hint 2
Find the value of $\lim _{n\rightarrow \infty }a_{n}$ to help you evaluate the limit $\lim _{n\rightarrow \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|$ appearing in the ratio test.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We will apply the ratio test to determine the radius of convergence of the power series $\sum _{n=0}^{\infty }a_{n}x^{n}$. The ratio test says that the power series converges when $\lim _{n\rightarrow \infty }\left|{\frac {a_{n+1}x^{n+1}}{a_{n}x^{n}}}\right|=\lim _{n\rightarrow \infty }|x|\left|{\frac {a_{n+1}}{a_{n}}}\right|=|x|\lim _{n\rightarrow \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|<1.$ By the solution to the Question 6(a), $\lim _{n\rightarrow \infty }a_{n}=1.$ Therefore, since each $a_{n}$ is positive, $\lim _{n\rightarrow \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|=\lim _{n\rightarrow \infty }{\frac {a_{n+1}}{a_{n}}}={\frac {\lim _{n\rightarrow \infty }a_{n+1}}{\lim _{n\rightarrow \infty }a_{n}}}={\frac {1}{1}}=1.$ Hence, the series converges when $|x|<1.$ This shows that its radius of convergence is ${\color {blue}1}$.