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Find the radius of convergence of the power series $\sum\limits_n \frac{(-1)^nx^n}{\ln(n+1)}$

$\displaystyle R = \frac{1}{\limsup\limits_{n \to \infty}\sqrt[n]{ \bigg|\frac{(-1)^n}{\ln(n+1)}}\bigg|} = \limsup\limits_{n \to \infty}\ln(n+1)^{1/n}. $ But $1 \le \ln(n+1)^{1/n} \le (n+1)^{1/n}$

So $\limsup\limits_{n \to \infty} 1 \le \limsup\limits_{n \to \infty} \ln(n+1)^{1/n} \le \limsup\limits_{n \to \infty}\lim(n+1)^{1/n} = 1.$ So $R = 1$.

Is this correct?

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  • $\begingroup$ How do you pull out $\limsup$ from the denominator and it remains $\limsup$? $\endgroup$
    – A.Γ.
    Dec 31 '15 at 1:20
  • $\begingroup$ @A.G. $\frac{1}{\limsup\limits_{n \to \infty}\sqrt[n]{ \bigg|\frac{(-1)^n}{\ln(n+1)}}\bigg|} = \frac{1}{\limsup\limits_{n \to \infty}{ \frac{1}{\ln(n+1)^{1/n}}}} = \frac{1}{{ \frac{1}{\limsup\limits_{n \to \infty} \ln(n+1)^{1/n}}}} = \limsup\limits_{n \to \infty}\ln(1+n)^{1/n}$ $\endgroup$
    – GGG
    Dec 31 '15 at 1:45
  • $\begingroup$ $\frac{1}{\limsup}=\liminf$. $\endgroup$
    – A.Γ.
    Dec 31 '15 at 2:54
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You can apply the ratio test. Find $L = \lim_{n \to \infty} | \frac{(-1)^{n+1} x^{n+1} \ln (n+1)}{(-1)^n x^n \ln (n+2)}| = |x| \lim_{n \to \infty} \frac{\ln (n+1)}{\ln (n+2)} $. Apply L'Hospital's rule to get $L = |x|$. Thus the series will absolutely converge if $|x| <1$.

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