Since the given power series contains the powers $x^n$, the radius of convergence is given by

At the right endpoint $x\=R\=1$, the given power series becomes $\mathrm{\Σ} a_n R^n\=\mathrm{\Σ} 1\/\ln \left(n\right)$, which diverges by part (3) of the Limit-Comparison test if the comparison series is taken as the divergent series $\sum _{n\=2}1\/\left(n-1\right)$, the harmonic series in disguise.

The relevant calculation that must be made is $c\=\underset{n\to \infty }{lim}\frac{a_n R^n}{1\/\left(n-1\right)}\=\infty$. Since the comparison series diverges, so too does the power series at the right endpoint $x\=1$.

At the left endpoint $x\=-R\=-1$, the given power series becomes the alternating series $\mathrm{\Σ} {\left(-1\right)}^n\/\ln \left(n\right)$, which converges conditionally by the Leibniz test, once it is noted that $1\/\ln \left(n\right)$ decreases monotonically to zero as $n\to \infty$.

Hence, the interval of convergence is $\left[-R\,R\right)\=\left[-1\,1\right)$.

The comparison series $\sum 1\/\left(n-1\right)$ is a disguised form of the divergent harmonic series. By part (3) of the Limit-Comparison test, since the comparison series diverges, so too does the power series at the right endpoint $x\=1$.

At the left endpoint $x\=-R\=-1$, the given power series becomes the alternating series $\mathrm{\Σ} {\left(-1\right)}^n\/\ln \left(n\right)$, which converges conditionally by the Leibniz test, once it is noted that $1\/\ln \left(n\right)$ decreases monotonically to zero as $n\to \infty$.

Hence, the interval of convergence is $\left[-R\,R\right)\=\left[-1\,1\right)$.