Test divergent series using a comparison test Test series for convergence / divergence using a comparison test:
$$\sum_{n=1}^\infty\frac{n^2+1}{n^3+2}$$
Now, If it would be $$\sum_{n=1}^\infty\frac{n^2+1}{n^3-2}$$ then I could compare it as greater or equal to 1/n series which diverges, but since it is + 2 in the denominator then I am not sure what to do?
 A: You don't need to compare starting at $1$. You have
$$
\sum_{n=2}^\infty\frac{n^2+1}{n^3+2}\geq\sum_{n=2}^\infty \frac{n^2}{n^3+n^3}=\frac12\,\sum_{n=2}^\infty\frac1n=\infty. 
$$
A: $n \ge 2$;
$\dfrac{n^2+1}{n^3+2}> \dfrac{n^2+1}{n^3+n}=$
$\dfrac{n^2+1}{n(n^2+1)}=\dfrac{1}{n}$.
Comparison test, harmonic series is divergent.
A: Since$$\lim_{n\to\infty}\frac{\frac{n^2+1}{n^3+2}}{\frac1n}=1,$$you have$$\frac{\frac{n^2+1}{n^3+2}}{\frac1n}\geqslant\frac12$$if $n$ is large enough. In other words,$$\frac{n^2+1}{n^3+2}\geqslant\frac1{2n}.$$Since the series $\sum_{n=1}^\infty\frac1{2n}$ diverges, your series diverges too.
A: You also have:
$$\sum_{n=2}^\infty\frac{n^2+1}{n^3+2}\geq\sum_{n=2}^\infty \frac{n^2+1}{2n^3}=\frac12 \left (\,\sum_{n=2}^\infty\frac1n+\sum_{n=2}^\infty\frac1 {n^³} \right )$$
Harmonic serie is divergent. 
Sum of a divergent series and a convergent series is divergent.
A: As often, the simplest way is using equivalents, because of the following result:

Let $\sum_n a_n, \sum_n b_n $ two series with positive terms, and suppose $a_n$ and $b_n$ are asymptotically equivalent. Then both series converge or both diverge.

Here, it is  particularly easy to apply, because a polynomial is asymptotically equivalent to its leading term. Therefore
$$\frac{n^2+1}{n^3+2}\sim_\infty \frac{n^2}{n^3}=\frac 1n$$
and the latter is the harmonic series, which diverges.
