Comparison test $\sum_{n=1}^{\infty} \frac{1}{2n+7}$

Question

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.

Answer 1

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.

Answer 2

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 $$ \frac{1}{2n+7} \times n \to \frac{1}{2} > 0. $$