Series $\sum_{n=0}^{\infty} (x-4)^n $ converges in x=1 and diverges in x=9. I can say that the convergence radius (R) is at least 3 and at most 5, thus 3<R<5. Is there a way I can tell it even more accurate?
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$\begingroup$ That series diverges when $x=1$. $\endgroup$ – José Carlos Santos Aug 18 '20 at 17:26
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$\begingroup$ it should have been R $\endgroup$ – Rikib1999 Aug 18 '20 at 17:36
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$\begingroup$ What has that to do with my comment? You claimed that your series converges when $x=1$, and I told you it does not. $\endgroup$ – José Carlos Santos Aug 18 '20 at 17:53
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$\begingroup$ aha, sorry, so what is then the radius? $\endgroup$ – Rikib1999 Aug 18 '20 at 18:01
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$\begingroup$ Please meaningful set a title. $\endgroup$ – Yves Daoust Aug 18 '20 at 18:40
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Probably you have some confusion. Given series $\sum_{n=1}^\infty (x-4)^n$.we know that this power series will converge for $x=4$(why!).
Now I am going to use ratio test you can also use root test. \begin{equation} |\frac{a_{n+1}}{a_n}|=|\frac{(x-4)^{n+1}}{(x-4)}|=|(x-4)|=L(say). \end{equation} Then series will converge if $L<1$. If $L=1$ then you cannot decide. If $L>1$ then series will diverge.
If $L<1$, then: $|x-4|<1 \Rightarrow 3<x<5$.
If $L=1$, then: $|x-4|=1 \Rightarrow x=3,5$. If $x=3$ then the series surely divergent also if $x=5$(By necessary condition of convergence of a series). \begin{equation} ROC= \frac{upper ~value-lower ~value}{2}=\frac{5-3}{2}=1. \end{equation}
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$\begingroup$ thank you very much :) now it is clear to me $\endgroup$ – Rikib1999 Aug 18 '20 at 19:47
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Hint:
It is well-known that the geometric series
$$\sum_{n=0}^\infty a_0r^n$$ converges to $$\frac{a_0}{1-r}$$ iff $$|r|<1$$ and that means that the radius of convergence is one.
answered Aug 18 '20 at 18:39
