Well, my problem is this : I need to check if this series is convergent :

Sigma of (1 to infinity) : $\displaystyle \frac{log(n)}{2n^3-1}$

now, I did the following thing:

$\displaystyle \frac{log(n)}{2n^3-1}$ < $\displaystyle \frac{log(n)}{2n^3}$ < $\displaystyle \frac{2n}{2n^3}$=$\displaystyle \frac{1}{n^2}$

*Wherever I wrote ' < ', it also works with ' <= '

(The last one is known to be convergent)

All is left for me to prove is this:

2n>=log(n)

I tried to prove it with induction... We haven't yet 'officially' learned logarithms, so this is not obvious to use.

Thanks