Nice problem Which is bigger $e^a$ and $a^3$ 
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
 A: 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,
$$
\begin{align}
a &= \sum_{n=1}^{\infty} \frac{1}{n!} \left\{\left(\frac{d}{dx}\right)^{n-1} e^{nx/3} \right\}_{x=0} \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \left\{\left(\frac{n}{3}\right)^{n-1} e^{nx/3} \right\}_{x=0} \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \left(\frac{n}{3}\right)^{n-1} \\
&= \sum_{n=0}^{\infty} \frac{1}{(n+1)!} \left(\frac{n+1}{3}\right)^{n}.
\end{align}
$$
A: The solution to $e^a=a^3$ is given by the Lambert W function:
$a=-3 \text{W} \left( - \frac{1}{3} \right)$
Now, take a look at the summation definition of the Lambert W function.  If you change your limits so they start at $n=1$ instead of $n=0$, then it is easy to see that your summation is equal to the value of $a$ given above.
A: well if e^a and a^3 are graphed
the results are a=1.857 and a=4.5364
lim a=>-inf e^a is 0 and lim a=>-inf a^3 is -inf
 lim a=>inf e^a/a^3
lim a=>inf e^a/3a^2
lim a=>inf e^a/6a
lim a>inf e^a/6 is inf by L'hopital rule
so e^a will be greater than a^3 from interval (-inf,1.857)u(4.53864,inf)
