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If I am given a radius of convergence $R$, how can I find a power series, which has this $R$ as radius of convergence?

My thoughts: If I turn the formula of the radius of convergence around I get $\limsup_{n \to \infty} \sqrt[n]{|a_n|} = \frac{1}{R}$. So I need to find a sequence which satisfies this condition. For simplicity I will choose a positive sequence (so I can drop the absolute value), and which is convergent (so I can drop the $\limsup$ for a $\lim$). So I need a sequence which $a_n \to \frac{1}{R^n}$. Let's choose $a_n := \frac{1}{R^n} + \frac{1}{n}$. I plugged in a few values and saw that $\lim_{n \to \infty} \sqrt[n]{|a_n|}$ is indeed $\frac{1}{R}$ but only for $R \ge 1$. I think this is because $(a_n)_n$ is only convergent for $R \ge 1$.

How can I find a sequence for the remaining values $0 \lt R \lt 1$? And if you can think of a better solution than mine, please share this also.

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  • $\begingroup$ What is the radius of convergence of $\sum_n x^n/3^n $? $\endgroup$ – Andrés E. Caicedo Jan 9 '18 at 17:29
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Note that $\sum_{n=0}^{\infty}z^n$ converges to $\frac1{1-z}$ for $|z|<1$ and diverges for $|z|\ge1$ . Hence $\sum_{n=0}^{\infty}\Big(\frac zR\Big)^n$ converges to $\frac1{1-\frac zR}$ for $|z/R|<1$ (or, $|z|<R$). Hence $$\sum_{n=0}^{\infty}\Big(\frac zR\Big)^n$$ has radius of convergence $R$.

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According to d'Alembert test any sequence $a_n$ with $lim_{n\to\infty}{{a_n}\over{a_{n+1}}}$ existed and less than 1 has a convergent power series $\Sigma_{n=1}^{\infty}a_n$. One such sequence is $a_n=({x\over R})^n$. There also more other convergence tests you can refer to.

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Consider $$ \sum_n (\frac {x-a}{R})^n$$ Now find the radius of convergence and you will find R. You may try the Ratio Test or the Root Test for this purpose.

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