How to check the convergence of $\sum_{n=1}^\infty\frac{\ln(n^2+n+1)}{n^{3/2}}$ There is an example of the Limit Comparison test on my textbook, and it finds the convergence of this series: $$\sum_{n=1}^\infty\frac{\ln(n^2+n+1)}{n^{3/2}}$$
It starts off with the limit
$$\lim_{n\to\infty}\frac{\frac{\ln(n^2+n+1)}{n^{3/2}}}{\frac{1}{n^{4/3}}}=$$
$$\lim_{n\to\infty}\frac{\ln(n^2+n+1)}{n^{1/6}}=$$Using De L'Hopital
$$\lim_{n\to\infty}\frac{(\ln(n^2+n+1))'}{(n^{1/6})'}=$$
$$\lim_{n\to\infty}\frac{6(2x+1)n^{5/6}}{n^2+n+1}=0$$
And since $\sum_{n=1}^\infty\frac{1}{n^{4/3}}<+\infty$, from the Limit Comparison test, $\sum_{n=1}^\infty\frac{\ln(n^2+n+1)}{n^{3/2}}<+\infty$
I can understand the whole process, except for the start (with the limit). Why is the series divided by $\frac{1}{n^{4/3}}$ ? This seems pretty random to me.
How do I choose what to divide the series with?
 A: It's not random. You have to find another convergent series to prove it convergent. Here a P series is chosen where p>1. You can choose other p series also where p>1
A: Combine comparison and integral tests. First notice that the numerator
$$
\log (x^2 +x+1) = 2 \log x + \log(1+\frac{1}{x} + \frac{1}{x^2}) \sim \log x 
$$
So we can compare the summand to $\frac{\log x}{x^\frac{3}{2}}$. For $x>x_0$ the function is monotone decreasing (check this), so we can use the integral test: 
$$
\int_{1}^{\infty} \frac{\log x dx}{x^\frac{3}{2}}
$$
Use IBP starting with $\int x^{-\frac{3}{2}}dx$, and you will see that the integral sonverges, hence the sum converges, hence the origial sum converges. 
A: as $n\to \infty$, $\log n$ is slower that $n^\alpha$ with any $\alpha>0$. Hence from the denominator, you know the series converges. Hence for a test, you want to compare it with a convergent series $\frac{1}{n^p},p>1$ while $p<\frac{3}{2}$, so that you still have a positive power of $n$ in the denominator after cancellation. 
You can understand it like the denominator $n^{\frac{3}{2}}$ gives up some power to cancel the increase of $log n$
A: Indeed,for any $\epsilon > 0$,there is an integer $N(\epsilon) >0$ s.t.
$$\log n \leq n^{\epsilon } ,\forall n \geq N(\epsilon )$$
And it is well known that, for any $\epsilon > 0$,
$$\sum_{1}^{\infty} \frac{1}{n^{1+\epsilon}} < + \infty$$
So, one can use the Cauchy's Criterion to check that 
$$\sum_{1}^{\infty} \frac{\log(n^{2}+n+1)}{n^{\frac{3}{2}}}$$
is absolutely convergent.
