Series' - Convergence - Limit of sequences 
Examine the following series' for convergence:
a)$\sum_{n=1}^\infty \frac{n^3\cdot 3^n}{n!}$,
b)$\sum_{n=1}^\infty\frac{n}{(-2n)^n}$,
c)$\sum_{n=1}^\infty \frac{n!}{n^n}$,
d)$\sum_{n=1}^\infty\frac{n^n}{n!+n^n}$.

I'm learning for exams and read about some tests, or whatever they are called, to determine if a sequence converges. The ones I read were direct comparison test, root test and ratio test. To be honest, I don't understand how to decide which to use on which sequence and wouldn't know how to apply them anyway. Never used them.
But I was thinking, when they ask to determine if a series is converging, wouldn't it be enough to try finding the limit of the sequence?
But I guess those 4 series' above don't have limits of sequence, or at least none that are easily to determine.
Anyway, my ideas so far:
a) No idea what test to use here. My first guess was to go with the ratio test:
$\frac{\frac{(n+1)^3\cdot 3^{(n+1)}}{(n+1)!}}{\frac{n^3\cdot 3^n}{n!}}=\frac{(n+1)^3\cdot 3}{n^3(n+1)}=\frac{(n+1)^2\cdot 3}{n^3}$?
My guess would that since the denominator has a power of 3 and the nominator only a power of 2, that it would converge to zero? Is that the correct argumentation? Is that even the correct test to use here?
To be honest, I'm basically lost at the rest of them. I feel most comfortable using the ratio test, but I can't seem to get anywhere with it with the other 3.
Any help or tips on how to appraoch this?
 A: the first :
$$e^x=\sum_{0}^{\infty }\frac{x^n}{n!}$$
derive three times and multiply by $x$ in each time
$$e^x(x^3+3x^2+x)=\sum_{n=1}^{\infty }\frac{n^3x^n}{n!}$$
use $x=3$
so it converges
The third:
use the Ratio Test
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!}=\frac{(n+1)n!}{(n+1)^{n+1}}\frac{n^n}{n!}=\frac{n^n}{(n+1)^n}=\frac{1}{(1+1/n)^n}$$
$$\lim_{n\rightarrow \infty }\frac{1}{(1+1/n)^n}=\frac{1}{e}<1$$
it converges
The Fourth:
$$\frac{n^n}{n!+n^n}=\frac{1}{1+\frac{n!}{n^n}}$$
A: The series  b) is absolutely convergent, since $$\left|\dfrac{n}{(-2n)^n}\right|=\dfrac{n}{2^n n^n}=\dfrac{1}{2^n}\cdot \dfrac{1}{n^{n-1}} \leqslant \dfrac{1}{2^n}.$$
For a) and c) d'Alembert's ratio test gives respectively
$$\dfrac{3^{n+1}(n+1)^3 }{(n+1)!} \cdot \dfrac{n!}{3^n n^3 }=\dfrac{3}{n+1}\cdot \left(\dfrac{n+1}{n}\right)^3 \underset{n\to \infty}{\to} 0<1,$$
$$\dfrac{(n+1)!}{(n+1)^{n+1}} \cdot \dfrac{n^{n}}{n!}  = \dfrac{n^n}{(n+1)^n}=\dfrac{1}{\left(1+\tfrac{1}{n} \right)^n}\underset{n\to \infty}{\to} \dfrac{1}{e}<1.$$
