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](https://mathoverflow.net/questions/84958/is-sum-limits-n0-infty-xn-sqrtn-positive).


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.



Alternating sum of square roots of binomial coefficients
Here's a proof of the positivity of
$$
c\_n(\alpha) := \sum\_{r=0}^n (-1)^r {n\choose r}^\alpha
$$
for all even $n$ and real $\alpha < 1$. It follows
(via M.Wildon's clever $F(x) F(-x)$ trick at mo.84958) that
$\sum\_{n=0}^\infty \phantom. x^n / n!^{\alpha} > 0$ for all $x \in\bf R$.
[**EDIT** fedja has meanwhile provided a very nice direct proof of
the positivity of $\sum\_{n=0}^\infty \phantom. x^n / n!^{\alpha}$.]


The key is to write $c\_n(\alpha)$ as a finite difference
$$
\sum\_{r=0}^n \phantom. (-1)^r {n\choose r} \cdot {n\choose r}^{\alpha - 1}
$$
and show that the Gamma interpolation
$$
\bigl(\Gamma(r+1)\Gamma(n-r+1) / n!\bigr)^{1-\alpha}
= n!^{\alpha-1} \exp\bigl((1-\alpha) (\log\Gamma(r+1) + \Gamma(n-r+1)\bigr)
$$
of ${n\choose r}^{\alpha - 1}$ has a positive $n$-th derivative
for all $r \in [0,n]$.


This in turn follows from the fact that the expansion of
$\log\Gamma(r+1) + \log\Gamma(n-r+1)$ in a Taylor series about $r = n/2$
has positive $(r - (n/2))^k$ coefficient for each $k=2,4,6,\ldots$.
[The coefficient vanishes for odd $k$ because
$\log\Gamma(r+1) + \log\Gamma(n-r+1)$ is an even function of $r-(n/2)$.]
Indeed the well-known formula
$$
\log \Gamma(x) = -\gamma x - \log x + \sum\_{j=1}^\infty
 \left[ \frac{x}{j} - \log \left( 1 + \frac{x}{j} \right) \right]
$$
shows that the $k$-th derivative of $\log\Gamma(x)$ is positive
for all $x>0$ and $k=2,4,6,\ldots$, because this is true for
$-\gamma x - \log x$ and for each term in the sum; explicitly
the derivative is $k! \phantom. \sum\_{j=0}^\infty (x+j)^{-k}$ which is
positive termwise. Therefore in the Taylor expansion
$$
\log \Gamma(r+1) = \log(n/2)! + \sum\_{k=1}^\infty \phantom. g\_k (r-(n/2))^k
$$
each of $g\_2,g\_4,g\_6,\ldots$ is even.
Since $\log\Gamma(r+1) + \log\Gamma(n-r+1)$ is
$$
2\log(n/2)!
 + 2 \Bigl( g\_2 (r-(n/2))^2 + g\_4 (r-(n/2))^4 + g\_6 (r-(n/2))^6 + \cdots\Bigr),
$$
the claim follows. [**EDIT** David Speyer notes that the convergence
of the Taylor series on $|r-(n/2)| \leq n/2$ requires justification,
and that the justification is easy because the $\Gamma(z)$ has no zeros
and poles only at $0,-1,-2,\ldots$ so the radius of convergence is $(n/2)+1$.]
Multiplying by $1 - \alpha$ and substituting into the exponential series,
we deduce that $(\Gamma(r+1) \Gamma(n-r+1))^{1-\alpha}$, too,
is a positive combination of even powers of $r-(n/2)$.


Now if a function $g$ has positive $n$-th derivative, then its
first finite difference
$$
g(x+1) - g(x) = \int\_x^{x+1} g'(y) dy
$$
has positive $(n-1)$-st derivative; repeating this argument $n$ times,
we find that the $n$-th finite difference is positive, and we're done.


