Science:Math Exam Resources/Courses/MATH101/April 2016/Question 06 (b)

MATH101 April 2016

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Question 06 (b)
Determine the interval of convergence for the power series $\displaystyle \sum _{n=1}^{\infty }{\frac {(x-1)^{n}}{\sqrt {n}}}$.

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
Start with the Ratio test.

Hint 2
Determine the convergence at the points $x$ where the corresponding $a_{n}$ satisfies $\lim _{n\rightarrow \infty }{\Big |}{\frac {a_{n+1}}{a_{n}}}{\Big |}=1$.

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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 first apply the Ratio Test with $a_{n}={\frac {(x-1)^{n}}{\sqrt {n}}}$. As ${\begin{aligned}\lim _{n\rightarrow \infty }{\Big |}{\frac {a_{n+1}}{a_{n}}}{\Big |}&=\lim _{n\rightarrow \infty }{\bigg |}{\frac {{(x-1)^{n+1}}/{\sqrt {n+1}}}{{(x-1)^{n}}/{\sqrt {n}}}}{\bigg |}\\&=\lim _{n\rightarrow \infty }{\bigg (}{\frac {\sqrt {n}}{\sqrt {n+1}}}|x-1|{\bigg )}\\&=\lim _{n\rightarrow \infty }{\bigg (}{\frac {1}{\sqrt {1+1/n}}}|x-1|{\bigg )}=|x-1|,\end{aligned}}$ the series converges for $|x-1|<1$ (in other words, for $0<x<2$) and diverges for $|x-1|>1$. When $x=2$, the series reduces to $\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n}}}$, which is a divergent $p$-series with $p={\frac {1}{2}}$. On the other hand, when $x=0$, the series reduces to $\sum _{n=1}^{\infty }{\frac {(-1)^{n}}{\sqrt {n}}}$, which converges by the Alternating Series Test. So the interval of convergence is $[0,2)$.