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

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- Dec 31st 2008, 03:56 AMcyclicA problem in analysis
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

- Jan 1st 2009, 07:39 PMNonCommAlg
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 $\displaystyle a_n \geq \frac{e}{(r \ln n)^n}.$ - Jan 2nd 2009, 02:18 AMcyclicThank you very much.
Your understand is correct. I am very sorry that I don't know how to post mathematical formula.

- Jan 2nd 2009, 02:33 AMNonCommAlg
- Jan 2nd 2009, 04:20 AMcyclicYou are wrong
$\displaystyle e^{e^x}=a_nx^n$ is a series

- Jan 2nd 2009, 04:30 AMNonCommAlg
- Jan 2nd 2009, 04:39 AMcyclicYes, that is correct.
I think that the problem should be very difficult, becauce in a Chinese math forum there is no one can solve it

- Jan 2nd 2009, 04:51 AMcyclicI post the problem again
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