$ \lim_{n \to \infty} \frac{(-1)^n(3-n)}{(3n-5)}$ and ... To find limits:
$(a) \lim_{n \to \infty} \frac{(-1)^n(3-n)}{(3n-5)}$
$(b) \lim_{n \to \infty} \frac{n^3}{n!}$

For the first one the sequence is oscillating so it does not converge.
For the 2nd one I used ratio test: Let $a_n = \frac{n^3}{n!} $, then $\frac{a_{n+1}}{a_n} = \frac{n! \times (n+1)^3}{n^3 \times (n+1)!} = (1+ \frac1n)^3 \times \frac{1}{1+n}$. Thus,
$|\frac{a_{n+1}}{a_n}| < 1$, by ratio test limit is $0$.
Is the solutions correct?
 A: If one sets $\displaystyle a_n=\frac{(-1)^n(3-n)}{(3n-5)}$, one gets
$$
\lim_{n \to \infty}a_{2n}=-\frac13\color{\red}{\ne}\frac13=\lim_{n \to \infty}a_{2n+1}
$$ thus $\left\{a_n\right\}_{\infty}$ does not converge.
Your second point is fine.
A: For the first one, just because it's oscillating doesn't mean it diverges. For example, let $A(n) = \frac{(-1)^n}{e^n}$. The limit as $A \to \infty = 0$ because the denominator increases without bound.
1) Let's first figure out if the non-oscillating terms converge. $S(n) = \frac{3-n}{3n-5}$. As $n \to \infty$, $S(n)$ approximates $\frac{-n}{3n} = -\frac13$. That means as $n \to \infty$, the entire sequence alternates between $\frac13$ and $-\frac13$. Diverges.
2) Perfect use of the ratio test. I just know automatically that factorial increases faster than exponential, which increases faster than polynomial, which increases faster than logarithmic.
A: Another way for the second is
$$\dfrac{n^3}{n!}<\dfrac{n}{n}\dfrac{n}{n-1}\dfrac{n}{n-2}\dfrac{1}{n-3}<1\dfrac{n}{\frac{n}{2}}\dfrac{n}{\frac{n}{2}}\dfrac{1}{\frac{n}{2}}=\dfrac{8}{n}<\varepsilon$$
for $n>6$.
