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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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 $R=3.$