series convergence. Dirichlet test $$\sum(-1)^{\lfloor\frac{n^3+n+1}{3n^2-1}\rfloor}\frac{\ln(n)}{n}$$
I thought about using Dirichlet test. $\ln(n)/n$ is a decreasing sequence that tends to 0 but I have problem with proving it. I also can
t tell anything about $(-1)^{\lfloor\frac{n^3+n+1}{3n^2-1}\rfloor}$. 

To prove that $\frac{\log(n)}{n}$ is decreasing, notice that:
$$\frac{\log(n+1)}{n+1}=\frac{\log n}{n+1}+\frac{\log\left(1+\frac{1}{n}\right)}{n+1}\leq \frac{\log n}{n+1}+\frac{1}{n(n+1)}$$
hence if $n> e,$
$$\frac{\log(n+1)}{n+1}< \log n\left(\frac{1}{n+1}+\frac{1}{n(n+1)}\right)=\frac{\log n}{n}.\tag{1}$$
To prove that
$$ (-1)^{\left\lfloor\frac{n^3+n+1}{3n^2-1}\right\rfloor}\tag{2} $$
has bounded partial sums, notice that:
$$ \frac{n^3+n+1}{3n^2-1} = \frac{n}{3} + \frac{4n+3}{9n^2-3} $$
and:
$$ \forall n\geq 3,\qquad 0<\frac{4n+3}{9n^2-3}<\frac{1}{5}, $$
so for any $n\geq 3$ we have that $(2)$ behaves like the sequence: 
$$ \ldots, -1,-1,-1,1,1,1,-1,-1,-1,1,1,1,\ldots \tag{3}$$
that obviously has bounded partial sums.
