Is $ \sum\limits_{n=0}^\infty x^n / \sqrt{n!} $ positive? Is $$ \sum_{n=0}^\infty {x^n \over \sqrt{n!}} > 0 $$ for all real $x$?
(I think it is.)  If so, how would one prove this?  (To confirm:  This is the power
series for $e^x$, except with the denominator replaced by $\sqrt{n!}$.)
 A: This comment serves to record a partial attempt, which didn't get very far but might be useful to others. Following a suggestion of Mark Wildon and Arthur B, define
$$f_n(\alpha) := \sum (-1)^r \binom{n}{r}^{\alpha}.$$
This is zero for $n$ odd, so we will assume $n$ is even from now on.
Mark Wildon shows that it would be enough to show that $f_n(1/2) \geq 0$ for all $n$. 
It is easy to see that $f_n(0) = 1$ and $f_n(1)=0$. Arthur B notes that, experimentally, $f_n(\alpha)$ appears to be decreasing on the interval $[0,1]$. If we could prove that $f_n$ was decreasing, that would of course show that $f_n(1/2) > f_n(1) =0$.
I had the idea to break this problem into two parts, each of which appears supported by numerical data:
1. Show that $f_n$ is convex on $[0,1]$.
2. Show that $f'_n(1) < 0$.
If we establish both of these, then clearly $f_n$ is decreasing.
I have made no progress on part 1, but here is most of a proof for part 2. We have
$$f'_n(1) = \sum (-1)^r \binom{n}{r} \log \binom{n}{r} = \sum (-1)^r \binom{n}{r} \left( \log(n!) - \log r!- \log (n-r)! \right)$$
$$=-2 \sum (-1)^r \binom{n}{r} \left( \log(1) + \log(2) + \cdots + \log (r) \right)$$ $$=-2 \sum (-1)^r \binom{n-1}{r} \log r.$$
At the first line break, we combined the $r!$ and the $(n-r)!$ terms (using that $n$ is even); at the second, we took partial differences once.
This last sum is evaluated asymptotically in this math.SE thread. The leading term is $\log \log n$, so the sum is positive for $n$ large, and $f'_n$ is negative, as desired. The sole gap in this argument is that the math.SE thread doesn't give explicit bounds, so this proof might only be right for large enough $n$.
This answer becomes much more interesting if someone can crack that convexity claim.
A: Here is a plot of
$$\frac{1}{100}\left(\sum_{k=0}^{16}\frac{x^k}{\sqrt{k!}}\right)$$
on the interval $[-4,0]$. (Above I added the terms up to degree $16$.)

Next, is a plot of 
$$\frac{1}{100}\left(\sum_{k=0}^{15}\frac{x^k}{\sqrt{k!}}\right)$$
on the interval $[-3,0]$.  (Above I added the  terms up to degree $15$)

This is one strange series.
A: Although the following does not provide another proof (perhaps it is possible to attempt one on this basis) I found it nice to see the following pictures.
Let's take from the series $f(x) = \sum_{k=0}^\infty {x^k \over \sqrt{k!}}$ the following variants in the same spirit as we have the hyperbolic and trigonometric series from the exponential-series:
$$\begin{array}{}
 \small \exp_{\tiny \sqrt{\,}}(x) &=& f(x) \\
 \small \cosh_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty {x^{2k} \over \sqrt{(2k)!}} \\
 \small \sinh_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty {x^{2k+1} \over \sqrt{(2k+1)!}} \\
\small  \tanh_{\tiny \sqrt{\,}}(x) &=& { \sinh_{\tiny \sqrt{\,}}(x)\over \cosh_{\tiny \sqrt{\,}}(x)  } \\
\small  \cos_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty (-1)^k {x^{2k} \over \sqrt{(2k)!}} \\
\small  \sin_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty (-1)^k {x^{2k+1} \over \sqrt{(2k+1)!}} \\
\end{array}$$ 
The answer to your question is equivalent to say, that always (="for real $x$")          


*

*$\small \cosh_{\tiny \sqrt{\,}}(x)$ is larger than $\small \sinh_{\tiny \sqrt{\,}}(x) $ $\qquad \qquad$ or that

*$\small \mid \tanh_{\tiny \sqrt{\,}}(x) \mid \lt 1$



To illustrate this I've plotted the $\sinh_{\tiny \sqrt{\,}}$ and $\cosh_{\tiny \sqrt{\,}}$-curves:              
 
This gives surely an extremely familiar impression...            
The $\tanh_{\tiny \sqrt{\,}}$-curve looks completely familiar too:
 
and the image suggests, that indeed the absolute value of $\small  \tanh_{\tiny \sqrt{\,}}(x) $ very likely is smaller than $1$ for all real $x$.

However, things are different for the $\sin_{\tiny \sqrt{\,}}$ and $\cos_{\tiny \sqrt{\,}}$ curves - they deviate strongly from the nicely periodic common trigonometric functions:                  

 
and combined they do not give a circle, but some ugly thing, strongly distorted (y-axis by $\small \cos_{\tiny \sqrt{ \,} }(\phi)$, x-axis by $\small \sin_{\tiny \sqrt{ \,} }(\phi)$, $\phi$ from $-5$ to $+5$) :             


A: The affirmative answer follows from my response to this related question.
EDIT. Noam Elkies gave a nicer and more general argument here.
A: Here is another non-answer. In "Asymptotic Methods in Analysis", chapter 6, de Bruijn proves that
$$S(s,n)=\frac{2}{\pi}\Gamma(s)(2ns\log 2n)^{-s}\left(\sin(\pi s)+O\left((\log n)^{-1}\right)\right)$$
where
$$S(s,n)= \sum_{k=0}^{2n} (-1)^k \binom{2n}{k}^s$$
for all $0\le s\le\frac{3}{2}$. So at least this explains things asymptotically.
A: Looks like the computers really spoiled us :) 
GH gave a perfectly valid answer already but the cheapest way to prove positivity is to write $\int_0^1(1-t^n)\log(\frac 1t)^{-3/2}\,\frac{dt}t=c\sqrt n$ with some positive $c$ (just note that the integral converges and the integrand is positive, and make the change of variable $t^n\to t$). Hence $\int_0^1 (f(x)-f(xt))\log(\frac 1t)^{-3/2}\,\frac{dt}t=cxf(x)$. If $x$ is the largest zero of $f$ (which must be negative), then plugging it in, we get $0$ on the right and a negative number on the left, which is a clear contradiction. Thus, crossing the $x$-axis is impossible. Of course, there is nothing sacred about $1/2$. Any power between $0$ and $1$ works just as well.
