Define convergence of series (criterion of Weierstrass)

Question

Using the criterion of Weierstrass define convergence of series: $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}n!}{n^{2n} }\cos2nx$$
$x\in R$

Answer 1

Find sequence $A_{n}$ which satisfy the inequality:

$\vert f_{n}\vert \le A_{n}$

$$\left|\frac {(-1)^{n+1}n!}{n^{2n} }\ \cos(2nx) \right|\le \frac{n!}{n^{2n}}$$

Now define the convergence of series: $\sum_{n=1}^{\infty}\frac{n!}{n^{2n}}$

Use D'Alambert's criterion:

$$\lim_{n \to \infty}\frac{A_{n+1}}{A_{n}} = \lim_{n \to \infty}\frac{(n+1)n!}{(n+1)^{2(n+1)}}\frac{n^{2n}}{n!} = \lim_{n \to \infty}\left(\frac{n}{n+1}\right)^{2n}\frac{1}{n+1}=0<1$$

$$ \left\{ \lim_{n \to \infty}(\frac{n}{n+1})^{2n} =\lim_{n \to \infty} \left(\frac{n+1-1}{n+1}\right)^{2n} = e^{-2} \right\} $$

Under D’Alambert criterion we find that $\sum_{n=1}^{\infty}\frac{n!}{n^{2n}}$ is convergent.

so

Under the criterion of Weierstrass we find that $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}n!}{n^{2n} }\ \cos (2nx)$ is uniformly convergent for $x\in R$.