# 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) : $\displaystyle \frac{log(n)}{2n^3-1}$

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

2. Originally Posted by adam63
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
You could consider the function

$\displaystyle f(x) = 2x - \ln x$
and show that $\displaystyle f(1) > 0$ and $\displaystyle f'(x) > 0$ for $\displaystyle x > 1$ so $\displaystyle f(x) > 0$ for all $\displaystyle x > 1$ (MVT).