# Series convergence radius proximity

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?

• That series diverges when $x=1$. – José Carlos Santos Aug 18 '20 at 17:26
• it should have been R – Rikib1999 Aug 18 '20 at 17:36
• What has that to do with my comment? You claimed that your series converges when $x=1$, and I told you it does not. – José Carlos Santos Aug 18 '20 at 17:53
• aha, sorry, so what is then the radius? – Rikib1999 Aug 18 '20 at 18:01
• Please meaningful set a title. – Yves Daoust Aug 18 '20 at 18:40

## 2 Answers

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.

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). $$\begin{equation} ROC= \frac{upper ~value-lower ~value}{2}=\frac{5-3}{2}=1. \end{equation}$$

• thank you very much :) now it is clear to me – Rikib1999 Aug 18 '20 at 19:47
• @Rikib16 you're welcome! – John Nash Aug 18 '20 at 19:49

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.

• thanks a lot! :) – Rikib1999 Aug 18 '20 at 19:47