Proving monotonic decreasing of general term: $ \sum_{n=2}^\infty (-1)^{n+1}\dfrac{n^2-1}{n^3-1} $

Question

I have got the following series -

$$ \sum_{n=2}^\infty (-1)^{n+1}\dfrac{n^2-1}{n^3-1} $$

I know that this an alternating series which converge - but got confused on how to prove this the general term series monotonic decreasing.

Should I use the derivate of it? or proving by $\frac{a_{n+1}}{a_{n}}$ as my teacher did ? and when should i choose each method?

Answer 1

Should I use the derivate of it? or proving by $\frac{a_{n+1}}{a_{n}}$ as my teacher did ?

You may use standard properties of precalculus, from $$ n+1\geq n $$ you get $$ (n+1)^2+(n+1)+1\geq n^2+n+1 $$ giving, for $n\geq1$,

$$ \frac1{(n+1)^2+(n+1)+1}\leq \frac1{n^2+n+1}. $$