Question: Determine whether the following series converge absolutely, converge conditionally or diverge

$$\sum\_{1}^{+ \infty}\frac{1}{n! + n}$$
I tried using the ratio test but i can seem to find the value of limit. For comparison test there seems to be no other suitable series to be compared with. Divergence test does not work either because limit will be zero

Answer: 
$$0 \leq \sum\_{n = 1}^{k}\frac{1}{n! + n} \leq \sum\_{n = 1}^{k}\frac{1}{n!}$$ and the series $\sum\_{n = 1}^{+ \infty}\frac{1}{n!}$ is convergent by the ratio test
For the ratio test (since all terms are positive, we drop the absolute values):
$$L = \lim\_{n\to\infty} \frac{n!+n}{(n+1)!+(n+1)} = \lim\_{n\to\infty} \frac{n!}{(n+1)!+n+1} + \lim\_{n\to\infty} \frac{n}{(n+1)!+n+1}$$
$$= \lim\_{n\to\infty} \frac{n!}{(n+1)!+n+1} + \lim\_{n\to\infty} \frac{n+1}{(n+1)!+n+1} - \lim\_{n\to\infty} \frac{1}{(n+1)!+n+1}$$
$$= \lim\_{n\to\infty} \frac{n!}{(n+1)!+n+1} + \lim\_{n\to\infty} \frac{1}{(n)!+1} - \lim\_{n\to\infty} \frac{1}{(n+1)!+n+1}$$
The second term goes to zero, and the third term goes to zero faster than the second term. So we focus on the first term. Factor out $n!$ from the denominator:
$$=\lim\_{n\to\infty} \frac{n!}{n!\left(n+1+\frac{n+1}{n!}\right)} =\lim\_{n\to\infty} \frac{1}{\left(n+1+\frac{n+1}{n!}\right)} =\lim\_{n\to\infty} \frac{1}{\left(n+1+\frac{1}{(n-1)!}+\frac{1}{n!}\right)}=0$$
So the series converges. But the comparison test is certainly an easier test to use here.