TITLE: Nice problem Which is bigger $e^a$ and $a^3$
QUESTION [7 upvotes]: let $$a=\sum_{n=0}^{\infty}\dfrac{\left(\dfrac{n+1}{3}\right)^n}{(n+1)!}$$

My Question:

Which is bigger $e^a$ and $a^3$

I guess 

$$e^a=a^3$$

but I can't prove it,and I think this is nice problem.Thank you evryone
This problem from this http://www.math.org.cn/forum.php?mod=viewthread&tid=3577&extra=&page=1

REPLY [7 votes]: We can "avoid" the Lambert $W$ function by skipping directly to its series representation.  We'll call upon the Lagrange inversion formula, which can be found in my answer here.
Suppose we're looking for the smallest positive root of the equation $e^x = x^3$, which is also the smallest positive root of the equation
$$
\frac{x}{e^{x/3}} = 1.
$$
Let's call this root $a$.  The Lagrange inversion formula yields the desired series representation,
$$
\begin{align}
a &= \sum_{n=1}^{\infty} \frac{1}{n!} \left\{\left(\frac{d}{dx}\right)^{n-1} e^{nx/3} \right\}_{x=0} \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \left\{\left(\frac{n}{3}\right)^{n-1} e^{nx/3} \right\}_{x=0} \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \left(\frac{n}{3}\right)^{n-1} \\
&= \sum_{n=0}^{\infty} \frac{1}{(n+1)!} \left(\frac{n+1}{3}\right)^{n}.
\end{align}
$$