Question

determine the radius and interval of convergence of the series:
as n=2 approaches infinity
(-1)^n x^n / (11^n ln(n) )

Answer 1

\[\sum_{n=2}^{infty}(-1)^n \frac{ x^n }{ 11^n \ln(n) }\]

Answer 2

\[\large\lim_{n\to\infty}\frac{(-1)^{n+1}}{11^{n+1}\ln(n+1)}\cdot\frac{11^{n}\ln(n)}{(-1)^{n}}=\lim_{n\to\infty}\left(-\frac{1}{11}\cdot\frac{\ln(n)}{\ln(n+1)}\right)=-\frac{1}{11}\]
The radius of convergence \(r\) is the reciprocal of this result. Since this series is centered about \(x=0\), the interval of convergence will be \((0-r,0+r).\)