Let $$ c_n = \sum_{r=0}^n (-1)^r \sqrt{\binom{n}{r}}. $$ It is clear that $c_n = 0$ if $n$ is odd. Remarkably, it appears that despite the huge positive and negative contributions in the sum defining $c_{2m}$, the sequence $(c_{2m})$ may be very well behaved.

Is $c_n > 0$ for all even $n$?

An affirmative answer will imply that the function $F(x) = \sum_{n=0}^\infty x^n/\sqrt{n!}$ is always strictly positive, thereby answering this earlier question.

Numerical computation using Magma shows that $c_n > 0$ if $n$ is even and $n \le 2000$. To give some illustrative values, $c_{100} = 0.077737 \ldots$, $c_{1000} = 0.019880 \ldots $ and $c_{2000} = 0.013317 \ldots$.

A comment by Mark Sapir on the earlier question suggests a stronger result might hold.

Is $c_{n} > c_{n+2} > 0$ for all even $n$?

I have checked that this is the case for all even $n \le 2000$.

It is very natural to ask what happens if we replace $\sqrt{\binom{n}{r}}$ with $\binom{n}{r}^\alpha$ for $\alpha \in (0,1)$. For $n\le 250$ the generalized version of the conjecture continues to hold if $\alpha = k/10$ where $k \in \mathbf{N}$ and $k \le 9$. Of course when $\alpha = 1$ we have $c_n = 0$ for all $n$, so, as David Speyer remarked in a comment on the earlier question, there is a good reason for the cancellation in this case.