Suppose e^(e^x)=an x^n, prove that an>=e(r ln n)^(-n) when n>1, r is a constant larger than e. Thanks
the reason that you haven't got any response is that your question is not very clear. i don't know about other members in here but if a poster didn't bother to explain his/her question clearly, i.e.
mathematically understandable, then i would just ignore the question. your question may be understood in different ways. for example, this one might be what you meant:
suppose $\displaystyle e^{e^x}=a_nx^n,$ for some $\displaystyle x \in \mathbb{R}, \ a_n \in \mathbb{R}, \ 1 < n \in \mathbb{N}.$ then there exists a real constant $\displaystyle r > e,$ independent from $\displaystyle n$ and $\displaystyle x,$ such that $\displaystyle a_n \geq \frac{e}{(r \ln n)^n}.$
Suppose $\displaystyle e^{e^x}=\sum_{n=0}^{\infty}a_nx^n$, prove that
$\displaystyle a_n \geq \frac{e}{(r \ln n)^n}$ when n>1, r is a constant larger than e.
Note that: NOt there exists a real constant r > e, BUT r is a constant larger than e