How can I test for convergence? The series is:
$$
\sum_{n=1}^{\infty} (-1)^{n + 1} \frac{3^n}{n! + 28}.
$$
I know we can say $a_n = 3^n/(n!+28)$ and use the alternating series test.
 A: using
$$\left(\frac{n}{e}\right)^n<n!$$
we have
$$\left|(-1)^{n + 1} \frac{3^n}{n! + 28}\right| \leqslant \frac{3^n}{n!}<\left(\frac{3e}{n}\right)^n$$
when we use power test for last we obtain
$$\sqrt[n]{\left(\frac{3e}{n}\right)^n} = \frac{3e}{n}\to 0$$
which gives absolute convergence, as I wrote above in comments.
A: We can do even better than just convergence and prove absolute convergence (which implies convergence). Indeed notice that
$$\lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_n}\right\rvert=\lim_{n\to\infty}\frac{\frac{3^{n+1}}{(n+1)!+28}}{\frac{3^n}{n!+28}}=\lim_{n\to\infty}3\cdot\frac{n!+28}{(n+1)!+28}=\lim_{n\to\infty}3\cdot\frac{1+\frac{28}{n!}}{n+1+\frac{28}{n!}}=0,$$
and so, by the ratio test, the series converges absolutely.
A: Consider the power series $\sum^\infty_{n=1}\frac{x^n}{n!}$, since
$$ \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}}=\frac{1}{n+1}\to 0,n\to \infty,$$
the radius of convergence of this power series is $\infty$. Hence $\sum^\infty_{n=1}\frac{3^n}{n!}$ converges. Since
$$ \frac{3^n}{n!+28}<\frac{3^n}{n!},$$
the series $\sum^\infty_{n=1}\frac{3^n}{n!+28}$ converges and thus $\sum_{n=1}^{\infty} (-1)^{n + 1} \frac{3^n}{n! + 28}$ converges absolutely.
