Test the convergence for this series $1+\frac{x}{2}+\frac{2!}{3^2}x^2+\frac{3!}{4^3}x^3+\ldots,$ for x > 0

Question

To study the convergence of $$1+\frac{x}{2}+\frac{2!}{3^2}x^2+\frac{3!}{4^3}x^3+\ldots,$$ I used ratio test and I got for $x/e < 1$ the series converges and for $x/e > 1$ series diverges and my doubt is if $x = e$ then whether the series will converge or diverge.

Answer 1

using stirling approximation.

https://en.wikipedia.org/wiki/Stirling%27s_approximation

$U_n = \frac{(n!e^n)}{(n+1)^n} = \sqrt{(2\pi n)} \frac{(\frac{n}{e})^ne^n}{(n+1)^n} $ = $\sqrt{(2\pi n)} \frac{n^n}{(n+1)^n} $

$ V_n = \frac{n^{n+0.5}}{n^n} = n^{0.5} $ Now by comparison test $\lim_{n \to \infty} U_n/Vn = \frac{\sqrt(2 \pi)}{e} $

this is a positive finite number so by comparison test as $V_n$ is divergent so $U_n$ is also divergent