This is the definition from Introduction to Algorithms (3rd edition) Chapter 3, page 63
For any monotonically increasing function $f(n)$ over the reals, for a given constant $c \in \mathbb{R}$, we define the iterated function $f_{c}^{\ast}(n)$ by
$$ f_{c}^{\ast}(n) = \min \left\{ i \geq 0: f^{(i)}(n) \leq c \right\} $$ where $f^{(i)}(n) = (\underbrace{f \circ f \circ ...\circ f}_{i})(n)$ with $ f^{(0)}(n) = n$.
Now the question asks for the bound (which should be tight as possible) on $f_{c}^{\ast}(n)$ for
$f(n) = n / \lg n$ where $c = 2$
I solved $f_{2}^{\ast}(n) = \omega(\sqrt{\lg n})$ (which means that $ \lim_{n \rightarrow \infty} f_{2}^{\ast}(n)/\sqrt{\lg n} = \infty$ ) and also solved $f_{2}^{\ast}(n) = O(\lg n)$ (which is pretty much evident). But my solution says that $f_{2}^{\ast}(n) = o(\lg n)$ (which means $\lim_{n \rightarrow \infty} f_{2}^{\ast}(n)/\lg n = 0$) which I can not justify mathematically. (I tried 1 day in this problem and can not obtain any progress.) Can someone give me some hints or scratch of proof?
