# Thread: Easy proof - Logarithms

1. ## Easy proof - Logarithms

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

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

now, I did the following thing:

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

*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

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

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

now, I did the following thing:

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

*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
You could consider the function

$
f(x) = 2x - \ln x
$

and show that $f(1) > 0$ and $f'(x) > 0$ for $x > 1$ so $f(x) > 0$ for all $x > 1$ (MVT).