Comparison test $\sum_{n=1}^{\infty} \frac{1}{2n+7}$ Comparison test $\sum_{n=1}^{\infty} \frac{1}{2n+7}$ 
My textbook did the limit comparison test to do this but cant you just do this using direct comparison test?
What I did
For $n \geq 1, \forall n \in \mathbb N$, $a_n = \frac{1}{2n+7} \leq \frac{1}{n} = b_n$
Since $$0 < a_n \leq b_n, \forall n \in \mathbb N$$
Direct comparison test applies
Consider $\sum_{n=1}^{\infty} \frac{1}{n}$. Since this is a p-series with magnitude one we know that $\sum b_n$ diverges. Therefore by the direct comparison test $\sum a_n$ diverges as well. 
Usually my textbook does the easiest way so I don't know. 
 A: Yes, you can use direct Comparison Test.
For all $n\geq 1$, we get
$$\frac{1}{2n+\color\red 7}\geq\frac{1}{2n+\color\red{7n}}=\frac{1}{9}\cdot\frac{1}{n}.$$
Since the series $$\sum_{n=1}^{\infty}\bigg(\frac{1}{9}\cdot\frac{1}{n}\bigg)$$ is a divergent series, it follows from the Comparison Test that the series $$\sum_{n=1}^{\infty}\frac{1}{2n+7}$$ is also divergent.
A: If you have got $b_{n} \leq a_{n}$, then $\sum_{n}b_{n}$ divergent implies $\sum_{n}a_{n}$ diverges. But you have got $b_{n} \geq a_{n}$; so there is no guarantee in this case. To apply comparison test, you have to seek another sequence $c_{n}$ ensuring that $a_{n} \geq c_{n}$ and that $\sum_{n}c_{n}$ diverges.
However, the limit comparison test can get you what you want immediately. Recall that the theorem states that if $a_{n},b_{n} \geq 0$ for all integers $n > 0$ and if $\lim_{n}a_{n}/b_{n}  > 0$ then either $\sum_{n}a_{n}$ and $\sum_{n}b_{n}$ both converges or both diverges. Note that
$$
\frac{1}{2n+7} \times n \to \frac{1}{2} > 0.
$$
