How can I determine the convergence of the following series So I've been solving convergence tests of series all day and I got stuck on the following three:


*

*$\sum_{k=0}^\infty \frac{4+|Cos{k}|}{k^3}$

*$\sum_{k=0}^\infty \frac{k!}{k^k}$

*$\sum_{k=0}^\infty \frac{1}{4+2^{-k}}$
For the first one I've tried to compare it to the series $\sum_{k=0}^\infty \frac{4}{k^3}$ which is a "smaller" series since |cosk| is always a number between 0 and 1. However that series seems to be converging since it is a "p" series where p=3 and p>1 and if the "smaller" series converges it doesn't have to mean that the larger series will converge as well. 
For the second one I've tried comparing it to the series $\sum_{k=0}^\infty \frac{k!}{k}$ but that didn't work as well and now I'm stuck at all three.
Can someone please give me an answer?
Any kind of help would be appreciated.
Thank you in advance!
 A: For the second series, note that
$$\frac{k!}{k^k}=\frac{k}{k}\,\frac{k-1}{k}\cdots \frac{2}{k}\,\frac 1k<\frac2{k^2}$$
and that the series $\sum_{k=1}^\infty \frac1{k^2}$ converges.
A: Let $a_k$ denote the $k$-th term.


*

*$|a_k| \leq \frac{5}{k^3}$.

*I'd use Stirling's approximation of the factorial.

*$a_k \geq \frac15$
A: 1) comparison tests
if  $\sum_\limits{k=1}^{\infty} a_k \le \sum_\limits{k=1}^{\infty} b_k$ and $\{b\}$ coverges, then $\{a\}$ converges
if  $\sum_\limits{k=1}^{\infty} a_k \ge \sum_\limits{k=1}^{\infty} b_k$ and $\{a\}$ diverges, then $\{b\}$ diverges.
Compare to $\sum\frac {5}{k^3}$ 
2) try the root test.
Update:
I was thinking to use Stirling's aproximation for $k!$  However, lets look at the ratio test, instead.
the series converges if $\lim_\limits{k\to\infty}\left(\dfrac {\frac {(k+1)!}{(k+1)^{k+1}}}{\frac {k!}{k^k}}\right)<1$
$\lim_\limits{k\to\infty}\frac {k^{k+1}}{(k+1)^{k+1}}\\
\lim_\limits{k\to\infty}(1-\frac {1}{k+1})^{k+1} = e^{-1} <1$
3) what is $\lim_\limits{k\to\infty} \frac{1}{4+2^{-k}}$?
