MHB Emil's question at Yahoo Answers (Radius of convergence)

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The discussion addresses Emil's question about finding the radius of convergence for the series ∑((x-3)^n)/(n3^n) from n=1 to infinity. The ratio test is applied, leading to the limit L = |x-3|/3. The series converges absolutely if |x-3| < 3 and diverges if |x-3| > 3. Consequently, the radius of convergence is determined to be R = 3. This mathematical explanation clarifies Emil's confusion regarding the concept.
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Hello Emil,

Using the ratio test to the series $\displaystyle\sum_{n=1}^{\infty}\dfrac{(x-3)^n}{n3^n}$ we get: $$L=\displaystyle\lim_{n \to\infty}\left |\dfrac{(x-3)^{n+1}}{(n+1)3^{n+1}}\cdot\dfrac{n3^n}{(x-3)^n}\right |=\dfrac{|x-3|}{3}\displaystyle\lim_{n \to\infty}\frac{n}{n+1}=\dfrac{|x-3|}{3}$$ If $\left |{x-3}\right |/3<1$ (or equivalently $\left |{x-3}\right |<3$) the series is absolutely convergent. If $\left |{x-3}\right |/3>1$ (or equivalently $\left |{x-3}\right |>3$) the series is divergent. As a consequence, the radius of convergence is $R=3.$
 
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