comparisons and logarithms

Hi. In comparisons, can work with logarithms in the usual manner? (Hoping, of course, my usual manner is the correct manner.)

Is this valid:

$\displaystyle \frac{ln(n+2)}{ln(n+1)} \le \frac{n+2}{n+1}$

$\displaystyle ln(n+2) \le \frac{ln(n+1)\,(n+2)}{n+1}$

$\displaystyle n+2 \le e^{\frac{ln(n+1)\,(n+2)}{n+1}}$

$\displaystyle n+2 \le (n+1)^{\frac{n+2}{n+1}}$

(Edit: assuming $\displaystyle n \ge 1$, if that matters.)

Re: comparisons and logarithms

Hey infraRed.

One approach you could try is showing d/dx [ln(x+2)] < d/dx (x+2) for all x >= 1 and show that ln(1+2) < 1+2.