We define a sequence, $\{a_n\}_{n=0}^\infty$, to be a uni-modal sequence if for some $m$, $$a_0<a_1<\cdots<a_m,\ \ \ \ a_m>a_{m+1}>a_{m+2}>\cdots.$$
We define a sequence, $\{a_n\}_{n=0}^\infty$, to be log-concave if for all $k$, $$\frac{a_k}{a_{k+1}}>\frac{a_{k-1}}{a_k}.$$
For every positive, uni-modal, log-concave sequence, does there exist an $n$ so that, $$\left(\sum_{k=0}^n \binom{n}{k} (a_{k+1}-a_k)(-1)^{n-k}\right)\cdot\left(\sum_{k=0}^n \binom{n}{k} (a_{k+1}+a_k)(-1)^{n-k}\right) <0?$$
Note, that if we define $$f(x):=\sum_{k=0}^\infty \frac{a_k}{k!}x^k,$$ then $$\left.\left(\frac{d^k}{dx^k}e^{-x}(f'(x)-f(x))\right)\right|_{x=0} = \left(\sum_{k=0}^n \binom{n}{k} (a_{k+1}-a_k)(-1)^{n-k}\right)$$ and $$\left.\left(\frac{d^k}{dx^k}e^{-x}(f'(x)+f(x))\right)\right|_{x=0} = \left(\sum_{k=0}^n \binom{n}{k} (a_{k+1}+a_k)(-1)^{n-k}\right)$$
Hence, our question pertains to the functions $e^{-x}(f'-f)$ and $e^{-x}(f'+f)$; can we prove that these functions must have at least one Taylor coefficient of opposite sign?
