I'd be grateful if someone could tell me how to obtain the convergence radius of the aforementioned power series. Or, by Cauchy Hadamard, the limit of $(n!/(n^n))^{(1/n)}$ as n approaches infinity. Thanks, Paul
I tried the quotient criteria, but since the quotient of two consecutive coefficients is not bounded by any value strictly smaller than 1, it didn't yield a result.
You can use Stirling $$ n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n$$ to evaluate $$ \lim_{n\to\infty} \left(\frac{n!}{n^n}\right)^{1/n} = \lim_{n\to\infty} (2\pi n)^{1/2n}e^{-1} = e^{-1}.$$
By "quotient critiria" you refer to d'Alembert's Ratio Test?
\begin{aligned} \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n} &= \lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n} \newline &= \lim_{n\rightarrow \infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} \newline &= \lim_{n\rightarrow \infty}\frac{(n+1)n!n^n}{n!(n+1)^{n+1}} \newline &= \lim_{n\rightarrow \infty}\frac{n^n}{(n+1)^n} \newline &= (\lim_{n\rightarrow \infty}(\frac{n+1}{n})^n)^{-1} \newline &= (\lim_{n\rightarrow \infty}(1+\frac{1}{n})^n)^{-1} \newline &= e^{-1}< 1 \end{aligned}
$\frac{a_{n}}{a_{n+1}}=\frac{n!\left(n+1\right)^{n+1}}{n^{n}\left(n+1\right)!}=\frac{\left(n+1\right)^{n}}{n^{n}}=\left(1+\frac{1}{n}\right)^{n}\to e$
so by the ratio test, the radius of convergence is $e$.
The Ratio Test is an easy to use tool for many of the radius of convergence problems that we meet. In also works very well for this somewhat harder problem. The details have already been posted.
If we want to use what you call Cauchy-Hadamard, otherwise known as the Root Test, we need to obtain information about $(n!/n^n)^{1/n}$. Stirling's Formula gives fairly precise information about the size of $(n!)^{1/n}$. It is an important estimate, and has many uses.
For our particular problem, we can work with a less precise estimate. Sums are more familiar than products, so we work with $\ln(n!)$. Note that $$\ln(n!)=\ln 1 + \ln 2 + \cdots +\ln n.$$ The sum on the right is well approximated by an integral. More precisely, $$\int_1^n \ln(x) dx < \ln 1 + \ln 2 + \cdots +\ln n < \int_2^{n+1} \ln(x)dx.$$ Integrate. After some manipulation of inequalities, we find that $(n!/n^n)^{1/n} \to e^{-1}$ as $n \to\infty$.
