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? 
 A: 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 $\alpha<-1$, 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. $\sqrt{n^3+n+1}<Cn^{3/2}$ for some $C$. Then you have $\sum \dfrac{\sqrt{n^3+n+1}}{n^4}<C\sum n^{-5/2}$, which converges. So look for $C$ such that $n^3+n+1<C^2n^3$ 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.
A: the inequality is false, just use limit comparison with the convergent 
$$ \sum \frac{1}{n^{5/2}} $$
A: $\sqrt{n^3+n+1} \leq \sqrt{2n^3}$, so:
$$\frac{\sqrt{n^3+n+1}}{n^4} \leq \frac{\sqrt{2n^3}}{n^4}=\frac{\sqrt{2}}{n^{5/2}}.$$
Since $\sum_{n=1}^\infty \frac{\sqrt{2}}{n^{5/2}}$ converges, by the comparison test our series converges as well.
A: As this is a series with positive terms, you may use equivalents: a polynomial function is asymptotically equivalent to its leading term, so
$$\frac{\sqrt{n^3+n+1}}{n^4}\sim_\infty\frac{\sqrt{n^3}}{n^4}=\frac 1{n^{5/2}},$$
which is a $p$-series ($p=5/2$)  converging to $\zeta(5/2)$.
A: Hint: use the ratio test with ${1\over {n^{5\over 2}}}$
${{\sqrt{n^3+n+1}\over n^4}}$
$={{n^{3\over 2}\over n^4}}
{{\sqrt{1+{1\over n^2}+{1\over n^3}}}}$
$={1\over {n^{5\over 2}}}{{\sqrt{1+{1\over n^2}+{1\over n^3}}}}$
