Upper and lower bounds for $\log(\frac{n}{n-1})$ If we have a value $x>1$ and a constant $c$ then by making $k=O(\log x)$ we have $$x\cdot (1-c)^k \leq 1$$ Now, if $c$ is not a constant, but a function such as 1/n, what is the smallest k for which $$x\cdot \left(1-\frac{1}{n}\right)^k \leq 1$$ ? Clearly an easy computation gives the value $$k \leq \frac{\log(1/x)}{\log (1-\frac{1}{n})} = \frac{\log x}{\log{\frac{n}{(n-1)}}}$$ The problem is that I have no intuition for the magnitude of the value $\log \frac{n}{(n-1)}$ nor for the value of $\log(1-1/n)$. Are there nice functions f(n) and g(n) that under-approximate and overapproximate the function $log(\frac{n}{n-1})$? In other words, whare are the easiest to manipulate and accurate functions f(n) and g(n) such that $$f(n) \leq \log(\frac{n}{n-1}) \leq g(n)$$ ? obs: x is a function that depends on n.

Use the bounds $$1-\frac{1}{x}\leq \log(x)\leq x-1$$ for all $x>0$. For $x=\frac{n}{n-1}$ these become $$\frac{1}{n}\leq \log\left(\frac{n}{n-1}\right)\leq \frac{1}{n-1}.$$