Math Help - rate of convergence

1. rate of convergence

I want to find the rate of convergence of

lim as n goes to infinity of (ln(n+1) - ln(n)), which equals
lim as n goes to infinity of ln((n+1)/n)

I said that ln[(n+1)/n) must be less than or equal to (n+n)/n^2 and it seems to work. Then, I think the rate if O(1/n).

However, I'm not sure how exactly to bound the ln function. Can someone show me a bound or why this works?

2. Originally Posted by PvtBillPilgrim
I want to find the rate of convergence of

lim as n goes to infinity of (ln(n+1) - ln(n)), which equals
lim as n goes to infinity of ln((n+1)/n)

I said that ln[(n+1)/n) must be less than or equal to (n+n)/n^2 and it seems to work. Then, I think the rate if O(1/n).

However, I'm not sure how exactly to bound the ln function. Can someone show me a bound or why this works?
Note that $\log (x+1) \leq x$ for $x>-1$ this means $\log y \leq y - 1$ for $y>0$.

Thus, $\log \frac{n+1}{n} \leq \frac{n+1}{n} - 1 = \frac{1}{n}$.