Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $e^{e^x}=\sum\limits_{n\geq0}a_nx^n$, prove that

$$a_n\geq e(\gamma\log n)^{-n}$$

for $n\geq2$, where $\gamma$ is some constant great than $e$.

share|improve this question
    
Can I ask which topics have recently been covered in your course or book? –  bgins Apr 5 '12 at 21:42
add comment

2 Answers 2

up vote 3 down vote accepted

I present three different approaches below, which may or may not appeal to you in helping you to formulate your own answer.

I think you can get this from the Taylor series at $0$ just by differentiating. If $y=y^{(0)}$ is your function and $y^{(n)}=yf_n(e^x)$, then $f_0(t)=1$, $y'=ye^x\implies f_1(t)=t$ and $y^{(n+1)}=y'f_n(e^x)+y\,e^xf_n'(e^x)$ $\implies$ $f_{n+1}(t)=f_1f_n+t\,f_n'$. Now $a_n=\frac{e\,f_n(1)}{n!}$ and $f_n(1)=\href{http://en.wikipedia.org/wiki/Bell_number}{B_n}$ (as @Autolatry points out) and it shouldn't be hard to show that this sequence is increasing and has (more than) the desired growth.


Upon further reflection, we want to show that for $n\ge2$, $$ a_n \ge e \left( \gamma \, \log n \right)^{-n} $$ $$ \left( \gamma \, \log n \right)^{-n} \le \frac{a_n}{e} = \frac{B_n}{n!} $$ $$ \gamma \, \log n \ge \left( \frac{a_n}{e} \right)^{-1/n} = \left( \frac{n!}{B_n} \right)^{1/n} $$ $$ \gamma \ge \sup \gamma_n \qquad\text{for}\qquad \gamma_n = \frac1{\log n} \left( \frac{a_n}{e} \right)^{-1/n} = \frac1{\log n}\left( \frac{n!}{B_n} \right)^{1/n} $$

Experiment suggests that $\gamma_n$ has a global minimum at $\gamma_{37}=0.56352\,15372\,44847$, lies below its pseudonym, the Euler-Mascheroni constant, $0.57721\,56649\,01532\cdots$, for $17\le n\le 114$, and may have its global maximum at $\gamma_2=1.44269\,50408\,88963$, depending on its asymptotic value. Using a recent bound (Berend-Tassa 2010) for $B_n$ and Stirling's formula for the factorial, $$ \gamma_n \ge \frac{\log(n+1)}{0.792\,n\,\log n}\Bigl( n! \Bigr)^{1/n} \approx \frac{\log(n+1)}{0.792\,\,n\log n}\cdot\frac{n}{e}\cdot\left(2\pi n\right)^{1/2n} \rightarrow\frac1{0.792\,e}\approx0.46449\,, $$ so that $\gamma=\gamma_2=\sup_{n\ge2}\gamma_n$ is in fact the best constant we can choose.


If we start from $y=e^{e^x}=\sum_{n=0}^{\infty}a_n\,x^n$ and differentiate to get $\sum_{n=0}^{\infty}(n+1)a_{n+1}\,x^n=y'=e^x\cdot e^{e^x}=\left(\sum_{n=0}^{\infty}\frac{x^n}{n!}\right)\left(\sum_{n=0}^{\infty}a_n\,x^n\right)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^n\frac{a_k}{(n-k)!}\right)x^n$, then we need to show (perhaps inductively) that the recursion $(n+1)a_{n+1}=\sum_{k=0}^n\frac{a_k}{(n-k)!}$ implies the desired inequality. Now $a_0=a_1=e\cdot1$ and the first few values we care about are $a_2=e\cdot\frac22$, $a_3=e\cdot\frac56$, $a_4=e\cdot\frac{15}{24}$, which as we already know satisfy our inequality for $\gamma\ge\gamma_1$ for our inductive base ($n=2$). As inductive hypothesis (with cumulative induction), we assume the inequality for $k\le n$: $$ a_k \ge e \left( \gamma \, \log k \right)^{-k} $$ from which it follows by induction (and from the recursion) that $$ \eqalign{ \frac{a_{n+1}}{e}\,\left(\gamma\log(n+1)\right)^{n+1} &\ge \frac{\left(\gamma\log(n+1)\right)^{n+1}}{n+1} \sum_{k=0}^n \frac{(\gamma\log k)^{-k}}{(n-k)!} \\ & > \frac{\left(\gamma\log(n+1)\right)^{n+1}}{n+1} \sum_{k=0}^n \frac{(\gamma\log n)^{-k}}{(n-k)!} \\ & > \frac{\gamma\log(n+1)}{n+1} \sum_{k=0}^n \frac{(\gamma\log n)^{n-k}}{(n-k)!} \\ & > \frac{\gamma\log(n+1)}{n+1} \sum_{k=0}^n \frac{(\gamma\log n)^k}{k!} \\ & \color{red}{>} \frac{\gamma\log(n+1)}{n+1} \left(e^{\gamma\log n} - \frac{(\gamma\log n)^{n+1}}{(n+1)!} \right) \\ & = \frac{\gamma\log(n+1)}{n+1} \left(n^\gamma - \frac{(\gamma\log n)^{n+1}}{(n+1)!} \right) \\ & \color{blue}{\ge 1 \qquad\text{(what we want!)}} } $$ where the last inequality (in red) follows from the error theorem for Taylor series. On the subsequent line, the two terms in parentheses exhibit opposite asymptotic behavior: the former grows toward $\infty$ for all $\gamma > 1$, while the latter decays toward $0$. We need only show that the last expression is $\color{blue}{\ge1}$ for some fixed $\gamma$ and all $n\ge2$. Clearly, this is true asymptotically for all $\gamma > 1$ by the relative asymptotic growth of $\log n$ versus $n^{\gamma-1}$ (e.g. using L'Hopital's rule). I suspect a slight modification of this argument would yield the result more elegantly.

share|improve this answer
    
Due to the expansion of the function $e^{e^x}$, we know that $a_n=\frac{eB_n}{n!}$. –  Riemann Apr 5 '12 at 13:51
    
Yes, of course. Thanks! –  bgins Apr 5 '12 at 14:41
add comment

Hint:

\begin{equation} e^{e^{x}} = e \sum_{k=0}^{\infty} \frac{x^{k}B_{k}}{k!} \end{equation}

Where $B_{k}$ is the $k$-th Bell number which is the number of partitions of a set with k entries, or the number of equivalence relations on it. Starting with $B_{0} = B_{1} = 1$, the first few Bell numbers are: $$1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975 \ldots$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.