What is correct approach for test of divergence/convergence for $\sum_{n=1}^{\infty} \frac{\sqrt{n^3+n+1}}{n^4}$ What is the correct approach for determining if the following series is convergent or divergent? 

$$\sum_{n=1}^{\infty} \frac{\sqrt{n^3+n+1}}{n^4}$$

My thought process was to use the limit comparison test where: 

$$\sum_{n=1}^{\infty} \frac{\sqrt{n^3+n+1}}{n^4} \leq \frac{1}{n^4}$$

According the p-series test this series converges. Is this correct?
After some experience with such problems, the answer is obvious: the terms approximate $n^{-5/2}$, so the series converges. But how do we approach the problem from first principles?

The first step is to make a crude estimate of the magnitudes involved. In this case, we have numerator $\sqrt{n^3+n+1}\approx n^{3/2}$, and denominator $n^4$. This gives a crude estimate of $\dfrac{n^{3/2}}{n^4}=n^{-5/2}$.

As I expect you know, $\sum n^\alpha$ converges if $\alpha&lt;-1$, so this crude estimate converges.

The next step is to show that your crude estimate of the numerator is a good enough approximation. That means that it is less than the true value by a constant factor $C$, i.e. $\sqrt{n^3+n+1}&lt;Cn^{3/2}$ for some $C$. Then you have $\sum \dfrac{\sqrt{n^3+n+1}}{n^4}&lt;C\sum n^{-5/2}$, which converges. So look for $C$ such that $n^3+n+1&lt;C^2n^3$ for all $n$, and you are done. In fact you only need to find $C$ such that this inequality holds for all sufficiently large $n$, which often makes the problem much simpler.

In the previous paragraph, I implicitly used the fact that all the terms are positive. If this is not the case, you might have a little more work to do.
Hint: use the ratio test with ${1\over {n^{5\over 2}}}$

${{\sqrt{n^3+n+1}\over n^4}}$

$={{n^{3\over 2}\over n^4}}
{{\sqrt{1+{1\over n^2}+{1\over n^3}}}}$

$={1\over {n^{5\over 2}}}{{\sqrt{1+{1\over n^2}+{1\over n^3}}}}$