MATH101 April 2013
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Question 11

FullSolution Problems. In questions 4–12, justify your answers and show all your work. If
a box is provided, write your final answer there. Unless otherwise indicated, simplification of numerical answers is required in these questions.
Consider the power series $\displaystyle \sum _{n=1}^{\infty }{\frac {(1)^{n}(x+2)^{n}}{\sqrt {n}}}$, where $\displaystyle x$ is a real number. Find the interval of convergence of this series.

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 by finding the radius of convergence.

Hint 2

Use the Ratio Test to calculate the radius of convergence.

Hint 3

Remember to check the endpoints of the interval!

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Solution

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We begin by calculating the radius of convergence, using the Ratio Test:
 $\displaystyle {\begin{aligned}\lim _{n\to \infty }\left{\frac {(1)^{n+1}(x+2)^{n+1}}{\sqrt {n+1}}}\cdot {\frac {\sqrt {n}}{(1)^{n}(x+2)^{n}}}\right&=\lim _{n\to \infty }{\sqrt {\frac {n}{n+1}}}\cdot x+2\\&=\lim _{n\to \infty }{\sqrt {\frac {1}{1+{\tfrac {1}{n}}}}}\cdot x+2\\&=x+2\end{aligned}}$
By the Ratio Test, the series converges if $\displaystyle x+2<1$ and diverges if $\displaystyle x+2>1$. Now we need to check the endpoints.
When $\displaystyle x+2=1$, that is, $\displaystyle x=3$, we have
 $\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {(1)^{n}(x+2)^{n}}{\sqrt {n}}}&=\sum _{n=1}^{\infty }{\frac {(1)^{n}(1)^{n}}{\sqrt {n}}}=\sum _{n=1}^{\infty }{\frac {1}{\sqrt {n}}}\end{aligned}}$
which diverges by the pseries test, since $\displaystyle 1/2<1$.
When $\displaystyle x+2=1$, that is, $\displaystyle x=1$, we have
 $\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {(1)^{n}(x+2)^{n}}{\sqrt {n}}}&=\sum _{n=1}^{\infty }{\frac {(1)^{n}}{\sqrt {n}}}\end{aligned}}$
This is an alternating series with
 $\displaystyle {\begin{aligned}b_{n}&={\frac {1}{\sqrt {n}}}\end{aligned}}$
Clearly, $\displaystyle b_{n+1}\leq b_{n}$ and $\displaystyle \lim _{n\to \infty }b_{n}=0$. By the Alternating Series Test,
 $\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {(1)^{n}(x+2)^{n}}{\sqrt {n}}}&=\sum _{n=1}^{\infty }{\frac {(1)^{n}}{\sqrt {n}}}\end{aligned}}$
converges. Therefore, the interval of convergence is $\displaystyle (3,1]$.

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