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Determine the radius of convergence of the following power series.

$$ \sum_{n=0}^{\infty}{a_n}(x-2017)^n\: \text{ with }\: a_n = \begin{cases} 1/2\:\text{ if $n$ is even} \\ 1/3\:\text{ if $n$ is odd} \end{cases} \tag{1}\label{1} $$

As far as I know.... to find the radius of convergence it must be true that $$ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| < 1\tag{2}\label{2} $$

But now I'm stuck here ... assuming that \eqref{2} holds, how can I find the radius of convergence?

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    $\begingroup$ Should be $\lim_{n\to\infty}|a_{n}/a_{n+1}|$. If this limit exists then it's the radius of convergence. But this is not the case for this series, thus you need to find another way to determine the radius of convergence. $\endgroup$ – Frank Lu Apr 30 '18 at 6:09
  • $\begingroup$ i can 't able to find the another way .. can u help me ??@FrankLu $\endgroup$ – user476275 Apr 30 '18 at 6:12
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    $\begingroup$ You may apply the Cauchy-Hadamard theorem, the radius of convergence is given by $$R=\frac{1}{\limsup_{n\to\infty}|a_n|^{\frac{1}{n}}}.$$In this case $R=1$. $\endgroup$ – Frank Lu Apr 30 '18 at 6:13
  • $\begingroup$ thanks and pliz see my answer @FrankLu..as i have written the answer.....is it correct or not? $\endgroup$ – user476275 Apr 30 '18 at 6:25
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Frank Lu has already answered almost comprehensively in the comments to this question: however I think it is a nice idea to precise some points. The standard way of calculating the radius of convergence of a power series is perhaps by using the Cauchy-Hadamard formula ([1], remark 5.2, p. 517)

$$ R=\left(\limsup_{n\to\infty} \sqrt[n]{|a_n|}\right)^{-1}\tag{I}\label{I} $$

Due to the properties of $\limsup$, the value of $R$ calculated by formula \eqref{I} always exists.

The formula for $R$ involving the inverse limit quotient as calculated for the ratio convergence test for the series $\sum a_n$, i.e.

$$ R=\lim_{n\to\infty}\frac{|a_n|}{|a_{n+1}|}\tag{II}\label{II} $$

works only if the ratio ${|a_n|}/{|a_{n+1}|}$ converges ([1], theorem 5.4, p. 518), and in this case obviously gives the same value as calculated by \eqref{I}. In the question posed by the OP, limit \eqref{II} does not exists since the sequence is oscillating, therefore the correct answer should be calculated only by formula \eqref{I}, i.e. $$ R=\left(\limsup_{n\to\infty} \sqrt[n]{|a_n|}\right)^{-1}=1 $$

[1] Emanuel Fisher (1983), "Intermediate Real Analysis", Springer Verlag.

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  • $\begingroup$ Thanks @Daniele...but pliz edits ur answer,, $\endgroup$ – user476275 Apr 30 '18 at 7:13
  • $\begingroup$ @Michael. Thanks for accepting my answer: I am not able to see why there are problems in its screen rendering. I already edited it two times without succeeding in solving the issue: in the edit screen I correctly see all formulas, but when I save it, the first two show up only as LaTEX/MathML code. Perhaps some sys admin could help us. $\endgroup$ – Daniele Tampieri Apr 30 '18 at 7:43
  • $\begingroup$ @Michael: I succede in correcting the strange behavior. Apparently the tags and labels I placed to identify equations interact with the ones I added when I edited the text of your question. $\endgroup$ – Daniele Tampieri Apr 30 '18 at 9:17
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The series can be decomposed in two convergent parts,

$$\frac12\sum_{n=0}^\infty(x-2017)^{2n}+\frac13\sum_{n=0}^\infty(x-2017)^{2n+1}$$

and both have radius of convergence $1$. For $x=2016$ and $x=2018$, they diverge and so does the sum (there is no cancellation).

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  • $\begingroup$ @ Yves im not getting ...as both have radius of convergence 1 that mean Radius of convergence of given series must be 1+1 =2 $\endgroup$ – user476275 Apr 30 '18 at 6:52
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    $\begingroup$ @Michael: what's your rationale for summing the radii of convergence ? (other than random guess...) $\endgroup$ – Yves Daoust Apr 30 '18 at 7:20

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