# Radius of convergence of $\sum_n \frac{(-1)^nx^n}{\ln(n+1)}$

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?

• How do you pull out $\limsup$ from the denominator and it remains $\limsup$?
– A.Γ.
Dec 31 '15 at 1:20
• @A.G. $\frac{1}{\limsup\limits_{n \to \infty}\sqrt[n]{ \bigg|\frac{(-1)^n}{\ln(n+1)}}\bigg|} = \frac{1}{\limsup\limits_{n \to \infty}{ \frac{1}{\ln(n+1)^{1/n}}}} = \frac{1}{{ \frac{1}{\limsup\limits_{n \to \infty} \ln(n+1)^{1/n}}}} = \limsup\limits_{n \to \infty}\ln(1+n)^{1/n}$
– GGG
Dec 31 '15 at 1:45
• $\frac{1}{\limsup}=\liminf$.
– A.Γ.
Dec 31 '15 at 2:54

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 $|x| <1$.