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Find both the radius of convergence and interval of convergence of the power series.

$$\sum_n\frac{(-1)^n\ln n}{\sqrt{n}}(x-2)^n$$

I believe that the radius of convergence is $R=1$. But I'm having a hard time with this entire question. the $\ln n$ is really throwing me off when I use the ratio test.

Michael Hardy
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jerry2144
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  • "the ln(n) is really throwing me off when I use the ratio test." This is odd since $\log(n+1)/\log(n)\to1$ hence, typically this kind of terms is NOT a problem in the ratio test. – Did Apr 28 '14 at 19:31

1 Answers1

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Denote $u_n(x)$ the given expression and since $$\frac{\ln(n+1)}{\ln n}=\frac{\ln n+\ln\left(1+\frac1n\right)}{\ln n}\xrightarrow{n\to\infty}1$$ then we see that $$\lim_{n\to\infty}\left|\frac{u_{n+1}(x)}{u_n(x)}\right|=|x-2|<1\iff 1<x<3$$ hence the radius of convergence is $R=1$.

  • If $x=3$ then by the Leibniz theorem the series is convergent
  • If $x=1$ the series is divergent since $$\frac{\ln n}{\sqrt n}\ge \frac{ 1}{\sqrt n}$$ hence the interval of convergence is $(1,3]$.