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I need to find a power series with the radius of convergence of $r = 3$. Since a power series is in the form $a_n x^n$, how does this relate to the radius of convergence?

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  • $\begingroup$ Consider the ratio test. $\endgroup$ – AmorFati Nov 16 '17 at 22:10
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What about $\displaystyle\sum_{n=0}\left(\dfrac{z}{3}\right)^{n}$?

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You can start with $$1+x+x^2+x^3+...=\frac{1}{1-x}$$ The radius of convergence of this series is $1$, meaning that if $x>1$ the formula is not valid, and the sum will diverge. Now in order to get this radius to increase, we replace every $x$ with $y=x/3$. What you get is a series with radius of convergence 1 for $y$ or $3$ for $x$. You can write then $a_n=3^{-n}$

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