Show $\sum_{n=0}^{\infty} \frac{n! a^n}{(n+1)^{n+1}}<\pi \sqrt{\frac{2}{e(e-a)}}$ Show $$\sum_{n=0}^{\infty} \frac{n! a^n}{(n+1)^{n+1}}<\pi \sqrt{\frac{2}{e(e-a)}}$$ where $a<e$.
This problem comes from the original version:
$$\int_0^{\infty} \frac{{\rm d}x}{e^x-ax}<\pi \sqrt{\frac{2}{e(e-a)}}(a<e),$$ which can deduce the present one, since 
\begin{align*} \int_0^{+\infty} \frac{{\rm d}x}{e^x-a x}&=\int_0^{+\infty} \frac{e^{-x}}{1-a xe^{-x}}{\rm d}x\\ &=\int_0^{+\infty} \sum_{n=0}^{\infty} a^nx^ne^{-(n+1)x}{\rm d}x\\ &= \sum_{n=0}^{\infty} \int_0^{+\infty}a^nx^ne^{-(n+1)x}{\rm d}x\\ &=\sum_{n=0}^{\infty}\frac{n!a^n}{(n+1)^{n+1}} \end{align*}
 A: A much tighter bound will be derived:
$$(1) \quad S(a):=\sum_{n=0}^\infty \frac{n! \, a^n}{(n+1)^{n+1}} \le \frac{\sqrt{2}\pi}{a}
\Big(\frac{1}{\sqrt{1-a/e}} - g(a)\Big) \quad ,$$
where
$$ g(a)=1 + 2 c \log \big(\frac{1+\sqrt{1-a/e}}{2}\big) , \quad c= \frac{\sqrt{2}\, e}{\pi} - 1$$
The bound of the OP follows from setting $g(a)=0$, which is allowable because for all $0\le a < e,\quad g(a)<0.$ However, this action makes the expression blow up at the origin, which is a still a bound, but only good as $a \to e.$  However, if the region in which in which one is most interested is $a \to e,$ then replace the offending denominator of $a$ with $e.$  The expression (1) is very accurate.  It is an exact match at $a=0$ and preserves the asymptote as $a \to e.$  For $a=e-1/10,$ the error is less than 1.2% . 
With an index shift $S(a)$ can be rewritten as 
$$ S(a) = \frac{1}{a} \sum_{n=1}^\infty \frac{n!}{n^{n+1}}\,a^n \, .$$
Via Stirling's formula one can derive
$$\frac{n!}{n^{n+1}}= \sqrt{2} \pi e^{-n} \Big( \frac{1}{2^{2n}} \binom{2n}{n} \Big) \Big( 1 + \frac{c}{n} + \frac{25}{1152n^2} -...\Big) \, , \quad c=\frac{5}{24}$$
I've chosen to write it this way because every knows the quantity in the first big parentheses ~ $1/\sqrt{n}.$  We can neglect the terms after the first if we make $c$ large enough, converting the asymptotics into a bound. We leave it undertermined, for now. Thus
$$ S(a) \le \Big(\sum_{n=0}^\infty x^n  \frac{1}{2^{2n}} \binom{2n}{n} + c \sum_{n=0}^\infty \frac{x^n}{n}  \frac{1}{2^{2n}} \binom{2n}{n} \Big)\, , \text{ where } x=a/e.$$  The sums are solvable in closed form, and we get the expression of (1), albeit with $c$ undetermined.  However, we know that $S(0) = 1$ (the usual 0^0=1 convention is followed, and it also follows from the integral form.) Using L'Hopital's rule for expressions looking like 0/0, we get the equation 
$$ 1 = \frac{1+c}{2} \sqrt{2} \,\frac{\pi}{e}, $$ which gives us that value for $c.$ This particular value of $c$ also satisfies
$$ \frac{n!}{n^{n+1}} \le \sqrt{2} \pi e^{-n} \Big( \frac{1}{2^{2n}} \binom{2n}{n} \Big) \Big( 1 + \frac{c}{n} \Big) $$ with equality for $n=1.$  Furthermore this $c$ is about 0.223657, which is only slightly larger than $5/24.$ 
