So I'm in Real Analysis and right now we are discussing series. It's very interesting. There is a problem on my homework that I want to make sure I used the comparison test correctly for. Keep in mind we haven't proved derivatives, so we can't use L'Hopital's rule or any of that fancy stuff. We can use only the series tests and rigorous definition of limits of sequences ($|a_n - L| < \epsilon $). So here's the problem :
Decide if the series $$\sum_{n=3}^\infty \log \left(\frac{n^2+n+1}{n^2}\right)$$ converges or diverges. If it converges, find its value.
Here's my work: take $a_n = \log \left(\frac{n^2+n+1}{n^2}\right)$ and $b_n = \frac{1}{n}$.
$$\log \left(\frac{n^2+n+1}{n^2}\right) > \frac{1}{n}$$ take both sides to the power $e$
$$\frac{n^2+n+1}{n^2} > e^\frac{1}{n} \iff n^2+n+1 > e^\frac{1}{n} \cdot n^2$$
And this is true because $1 \leq e^\frac{1}{n}\leq e$ and $n^2+n+1 > n^2$. And we know $\frac{1}n$ diverges and $b_n$ is less than $a_n$, so $a_n$ also diverges.
Q.E.D.
To show that $\sum_n \log \frac {n^2+n+1}{n^2}$ diverges by the most elementary means:
$$\text {We have }\quad a_n=\log \left(\frac {n^2+n+1}{n^2}\right)>\log \left( \frac {n^2+n}{n^2}\right)=\log \left( \frac {n+1}{n}\right).$$ $$\text {Therefore }\quad \sum_{n=1}^Na_n>\sum_{n=1}^N\log \left(\frac {n+1}{n}\right)=$$ $$=\log \prod_{n=1}^N\left(\frac {n+1}{n}\right)=\log \left(\frac {(N+1)!/1!}{N!}\right)=\log \;(N+1).$$