# Minimising $\sum \frac{(-1)^{a_n}}{n!}$

Let $$a_n$$ be a sequence of positive integers. Then I want to find the minimum possible value of $$\Bigg|\sum_{n=0}^\infty \frac{(-1)^{a_n}}{n!}\Bigg|$$

I know that the maximum absolute value is $$e$$ given where all of $$a_n$$ are odd or even, but I am unsure how to approach minimising the summation.

• A more interesting question is how to minimize $$\left|\sum^\infty_{k=1}\frac{(-1)^{a_k}}{k}\right|$$ – Szeto Feb 28 at 9:02
• @ParclyTaxel How did you derive? – Szeto Feb 28 at 9:15
• @ParclyTaxel The sum of other terms is a harmonic series... – Szeto Feb 28 at 9:17
• @Szeto Basically: the harmonic series is divergent. We can choose signs such that the series converges to 0. – Parcly Taxel Feb 28 at 9:21
• @ParclyTaxel I think you have to elaborate. – Szeto Feb 28 at 9:24

We can arrange for the first two terms to cancel out by making $$a_0,a_1$$ of opposite parities, say $$a_0=1,a_1=2$$ – their magnitudes are both 1. But the magnitude of the $$n=2$$ term is larger than the sum of magnitudes of all the other terms. It follows that the minimum magnitude $$m$$ of the whole sum is achieved by setting all terms after $$n=2$$ to be of opposite sign as that of the $$n=2$$ term (i.e. $$a_2$$ is of opposite parity to $$a_3,a_4,\dots$$), with $$m=\frac12-\sum_{n=3}^\infty\frac1{n!}=3-e$$ A possible $$(a_n)$$ achieving this $$m$$ would be $$1,2,3,4,6,8,10,12,\dots$$