# Asymptotics of $\frac{1}{\log(\frac{2^n}{2^n-1})}$

I am trying to understand the asymptotics of

$$$$f(n) = \frac{1}{\log(\frac{2^n}{2^n-1})}$$$$ In particular, is there some $$c \geq 1$$ such that $$f(n) = O(n^c)$$?

From this post, you can approximate $$\log(1+x)$$ with $$x$$ for little values of $$x$$. Hence,

$$f(n) \sim \frac{1}{\frac{1}{2^n-1}} = 2^n-1$$

Therefore, you can't find any constant $$c$$, such that $$f(n) = O(n^c)$$, as it is $$\Theta(2^n)$$.

• I doubt this result since we are only interested in very large values of n (greater than some $n_0$). I even think the expression is true for reasonably small values of $c$ (close to 1). Commented Nov 17, 2019 at 16:00
• @narekBojikian I didn't get your idea. My analysis is also base on the large values of $n$.
– OmG
Commented Nov 17, 2019 at 16:04
• I think that the answer is correct. As $n\rightarrow \infty$, $\frac{1}{2^n -1} \rightarrow 0$ - i.e. for large values of $n$ it is indeed true that $\log(1 + \frac{1}{2^n-1}) \simeq \frac{1}{2^n-1}$.
– Ryan
Commented Nov 17, 2019 at 17:23
• Oh sorry my fault. I get it now. Commented Nov 17, 2019 at 21:38