Six numeric series: convergence study

(1) $\sum\frac{n!}{n^n}$. Ratio:
$$\frac{u_{n+1}}{u_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\left(\frac{n}{n+1}\right)^n=\left(1+\frac1n\right)^{-n}\to\frac1e<1.$$
By d'Alembert, the series converges.

(2) $\sum\frac{n!}{a^n}$, $a>0$. Ratio:
$$\frac{u_{n+1}}{u_n}=\frac{(n+1)!}{a^{n+1}}\cdot\frac{a^n}{n!}=\frac{n+1}{a}\to+\infty.$$
For every fixed $a>0$, the ratio eventually exceeds $1$, so $u_n\not\to0$: the series diverges.

(3) $\sum\frac{2n}{n+2^n}$. For $n\ge1$,
$$0\le\frac{2n}{n+2^n}\le\frac{2n}{2^n}=2n\cdot2^{-n}.$$
The series $\sum 2n\cdot2^{-n}$ converges (ratio test: $(n+1)/2n\to1/2<1$). By comparison, the series converges.

(4) $\sum\frac{1}{n\ln n(\ln\ln n)^\beta}$. Use the substitution $u=\ln\ln t$; the integral $\int\frac{dt}{t\ln t(\ln\ln t)^\beta}=\int\frac{du}{u^\beta}$ converges iff $\beta>1$. Hence the series converges if $\beta>1$ and diverges if $\beta\le1$.

(5) $\sum\frac{n^{\ln n}}{(\ln n)^n}$. Write
$$n^{\ln n}=\exp((\ln n)^2),\qquad (\ln n)^n=\exp(n\ln\ln n),$$
so
$$u_n=\exp\bigl((\ln n)^2-n\ln\ln n\bigr).$$
Since $n\ln\ln n$ dominates $(\ln n)^2$, one has $u_n=o(n^{-2})$. By comparison with $\sum n^{-2}$, the series converges.

(6) $\sum\frac{\cos^2 n}{\sqrt n}$. Since $\cos^2 n\ge0$ and $\cos^2 n\ge\frac12$ along a positive density of integers (e.g. $n\equiv0\pmod{2\pi}$ approximation), one gets $\frac{\cos^2 n}{\sqrt n}\not\to0$ fast enough. More directly: $\cos^2 n=\frac{1+\cos 2n}{2}$, so
$$\frac{\cos^2 n}{\sqrt n}=\frac{1}{2\sqrt n}+\frac{\cos 2n}{2\sqrt n}.$$
The first term gives a divergent series ($\sum n^{-1/2}$), while the second converges by Dirichlet's test. Hence the series diverges (to $+\infty$).