Find radius of convergence of $\displaystyle \sum_{n=1}^{\infty}\frac{x^{6n+2}}{(1+\frac{1}{n})^{n^2}}$.
My try:
Using Cauchy's Root test for convergence $\displaystyle \lim_nsup \sqrt[n]{|a_n|}|(z-c)|<1$, we get
$\displaystyle \lim_nsup \sqrt[n]{\left|\frac{1}{\Big(1+\frac{1}{n}\Big)^{n^2}}\right|}\,|x^6|<1\implies \lim_nsup \frac{1}{\Big(1+\frac{1}{n}\Big)^{n}}|x|^6<1$. Hence $|x|^6<{e}$ or
the interval of convergence is $\displaystyle \Big(-e^\frac{1}{6},e^\frac{1}{6}\Big)$ and radius of convergence is $2e^\frac{1}{6}$. Am I correct ??
That the intervall of convergence is $\displaystyle \Big(-e^\frac{1}{6},e^\frac{1}{6}\Big)$, is correct. But the radius of convergence is $e^\frac{1}{6}$.