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 $\sqrt{n!}$.)

Looks like the computers really spoiled us :) GH gave a perfectly valid answer already but the cheapest way to prove positivity is to write $\int_0^1(1t^n)\log(\frac 1t)^{3/2}\,\frac{dt}t=c\sqrt n$ 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. 


The affirmative answer follows from my response to this related question. EDIT. Noam Elkies gave a nicer and more general argument here. 


Here is another nonanswer. 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. 


Additional data for Liviu's plots. I used Pari/GP with 1200 digits dec prec, documenting also the required number of terms after which the absolute values of the summands of the series decrease below 1e100. There seems to be no local minimum... $\small \begin{array}{rlr} & & \text{# of terms}\\ x & f(x) & \text{ required} \\ \hline \\ 1 & 0.438599896749 & 201 \\ 2 & 0.247539616819 & 201 \\ 3 & 0.162554775870 & 211 \\ 4 & 0.117399404501 & 257 \\ 5 & 0.0903120618145 & 304 \\ 6 & 0.0726061182760 & 354 \\ 7 & 0.0602796213492 & 407 \\ 8 & 0.0512783927864 & 464 \\ 9 & 0.0444561508357 & 525 \\ 10 & 0.0391295513879 & 589 \\ 11 & 0.0348689168813 & 658 \\ 12 & 0.0313919770798 & 730 \\ 13 & 0.0285063993737 & 808 \\ 14 & 0.0260770215882 & 889 \\ 15 & 0.0240063146159 & 976 \\ 16 & 0.0222222780410 & 1067 \\ 17 & 0.0206706877888 & 1162 \\ 18 & 0.0193099849974 & 1263 \\ 19 & 0.0181078191003 & 1369 \\ 20 & 0.0170386561852 & 1479 \\ 21 & 0.0160820905671 & 1595 \\ 22 & 0.0152216309789 & 1715 \\ 23 & 0.0144438135509 & 1841 \\ 24 & 0.0137375438980 & 1972 \\ 25 & 0.0130936024884 & 2108 \\ 26 & 0.0125042681404 & 2250 \\ 27 & 0.0119630281606 & 2396 \\ 28 & 0.0114643528377 & 2548 \\ 29 & 0.0110035182996 & 2705 \\ 30 & 0.0105764661081 & 2867 \\ 31 & 0.0101796910429 & 3035 \\ 32 & 0.00981015071575 & 3208 \\ 33 & 0.00946519223932 & 3386 \\ 34 & 0.00914249232841 & 3569 \\ 35 & 0.00884000806032 & 3758 \\ 36 & 0.00855593615550 & 3953 \\ 37 & 0.00828867911422 & 4152 \\ 38 & 0.00803681690505 & 4357 \\ 39 & 0.00779908317617 & 4567 \\ 40 & 0.00757434517200 & 4783 \\ 41 & 0.00736158670179 & 5004 \\ 42 & 0.00715989363457 & 5231 \\ 43 & 0.00696844149585 & 5462 \\ 44 & 0.00678648482039 & 5700 \\ 45 & 0.00661334797911 & 5942 \\ 46 & 0.00644841724806 & 6190 \\ 47 & 0.00629113392871 & 6444 \\ 48 & 0.00614098836080 & 6703 \\ 49 & 0.00599751469633 & 6967 \\ 50 & 0.00586028632445 & 7236 \\ 51 & 0.00572891185489 & 7511 \\ 52 & 0.00560303158255 & 7792 \\ 53 & 0.00548231436720 & 8078 \\ 54 & 0.00536645487311 & 8369 \\ 55 & 0.00525517112099 & 8666 \\ 56 & 0.00514820231209 & 8968 \\ 57 & 0.00504530688991 & 9275 \\ 58 & 0.00494626080983 & 9588 \\ 59 & 0.00485085599129 & 9907 \\ 60 & 0.00475889893049 & 10230 \\ 61 & 0.00467020945455 & 10560 \\ 62 & 0.00458461960073 & 10894 \\ 63 & 0.00450197260623 & 11234 \\ 64 & 0.00442212199624 & 11580 \\ 65 & 0.00434493075923 & 11931 \\ 66 & 0.00427027059992 & 12287 \\ 67 & 0.00419802126157 & 12649 \\ 68 & 0.00412806991028 & 13016 \\ 69 & 0.00406031057475 & 13388 \\ 70 & 0.00399464363573 & 13766 \end{array} $ 


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: I have made no progress on part 1, but here is most of a proof for part 2. We have $$f'_n(1) = \sum (1)^r \binom{n}{r} \log \binom{n}{r} = \sum (1)^r \binom{n}{r} \left( \log(n!)  \log r! \log (nr)! \right)$$ $$=2 \sum (1)^r \binom{n}{r} \left( \log(1) + \log(2) + \cdots + \log (r) \right)$$ $$=2 \sum (1)^r \binom{n1}{r} \log r.$$ At the first line break, we combined the $r!$ and the $(nr)!$ 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. 


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. 


Another "notyetanswer"... I've tried another idea. Assume the function f(x) is expressed by the following composition:
$$\small x' = \exp(x)1 $$
$$\small f(x) = g(x') = g(exp(x)1) $$
The idea is, that the unavoidable big "hump" in the partial sums, after which the sequence of partial sums begins to decrease, may be absorbed by the function $\small g(x)$  because $\small \exp(x) $ is really small for large negative x and x' is then very little above 1.
I did not yet arrive at a conclusive result; but the power series for $\small g(x) $ begins with the smooth looking form (and gives the partial sums for $\small x'=\exp(100)1 $): The the question is, for some large negative x, say $\small x=100 \qquad x'=exp(100)1 = 1+ \epsilon $ the series $\ g(x') $ converges to zero. Unfortunately  although we've translated the original problem to one with nice small numbers I don't see, how to really come nearer a solution, because the convergence of $\small g(1+\epsilon) $ is really slow  if it converges at all to a positive value... So this is not yet a solution, but perhaps a suggestion for a path to try... 


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.
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 (yaxis by $\small \cos_{\tiny \sqrt{ \,} }(\phi)$, xaxis by $\small \sin_{\tiny \sqrt{ \,} }(\phi)$, $\phi$ from $5$ to $+5$) : 

