Consider the alternating series $$S_n = \sum_{k=0}^{n} (-1)^k {n\choose k} a_k$$ where the weights $a_k$ are non-negative, bounded and monotonically decreasing ($a_{k+1} < a_k$). Can it be shown that $\lim_{n \to \infty} S_n = 0$ for any choice of $a_k$?
What I've worked out so far is that setting $a_k = 1$ gives the simple result of $0$. Then using Abel's test, we can show that $S_n$ converges, although it doesn't give the limiting value of the series.
One method I can think of is to use a polynomial interpolation of the weights $a_k$, and then make use of the property that $$\sum_{k=0}^{n} (-1)^k {n\choose k} k^l = 0, \quad 0 \leq l < n,$$ but I'm wondering if there is a more elementary proof.
The answer is negative.
Let $f(z)=\sum_{n=0}^\infty a_n z^n$ and $g(z)=\sum_{n=0}^\infty S_n z^n$; it's easy to see that $f(z)$ converges at least for $|z|<1$, and $g(z)$ converges at least for $|z|<1/2$; under the last condition we have $$g(z)=\frac1{1-z}f\left(\frac{-z}{1-z}\right).$$
Thus, if $S_n\to 0$, then $g(z)$ must be analytic at least on $|z|<1$, and our strategy is to break this condition by choosing $f$ suitably.
Let $a_{2n}=\lambda^n$ and $a_{2n+1}=\lambda^n\mu$ with $0<\lambda<\mu<1$. Then $$f(z)=\frac{1+\mu z}{1-\lambda z^2},\quad g(z)=\frac{1-(1+\mu)z}{(1-z)^2-\lambda z^2},$$ and if $\lambda\neq\mu^2$ then $g(z)$ has a pole at $z=1/(1+\sqrt\lambda)$, causing $|S_n|$ to grow like $(1+\sqrt\lambda)^n$.
Say, at $\lambda=1/4$ and $\mu=1/3$ we get $$S_n=\frac{3^n+5}{2^n\cdot 6}\underset{n\to\infty}{\longrightarrow}\infty.$$