Example 1   Does the following series converge? $$\sumn \frac{1}{n^2+n+1}$$ The series is similar to $\sumn 1/n^2$ which converges (use $p$ -test, for example). Next we compute the limit: $$\limn \frac{\frac{1}{n^2+n+1}}{\frac{1}{n^2}}=\limn \frac{n^2}{n^2+n+1} = 1$$ Therefore, since $1\neq 0$ , by the Limit Comparison Test (with $a_n=1/(n^2+n+1)$ and $b_n=1/n^2$ ), the series converges.

Example 2   Does the following series converge? $$\sumn \frac{n^3+n+1}{n^4+n+1}$$ If we ``forget'' about the lower order terms of $n$ : $$\frac{n^3+n+1}{n^4+n+1} \sim \frac{n^3}{n^4}=\frac{1}{n}$$ and $\sumn 1/n$ is the harmonic series which diverges (by the $p$ -test). Thus, we take $b_n=1/n$ and compute: $$\limn \frac{\frac{n^3+n+1}{n^4+n+1}}{\frac{1}{n}}=\limn \frac{n(n^3+n+1)}{n^4+n+1}= \limn \frac{n^4+n^2+n}{n^4+n+1}=\limn \frac{1+1/n^2+1/n^3}{1+1/n^3+1/n^4}=1$$ Therefore the series diverges like the harmonic does.