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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.

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  • $\begingroup$ A more interesting question is how to minimize $$\left|\sum^\infty_{k=1}\frac{(-1)^{a_k}}{k}\right|$$ $\endgroup$ – Szeto Feb 28 at 9:02
  • $\begingroup$ @ParclyTaxel How did you derive? $\endgroup$ – Szeto Feb 28 at 9:15
  • $\begingroup$ @ParclyTaxel The sum of other terms is a harmonic series... $\endgroup$ – Szeto Feb 28 at 9:17
  • $\begingroup$ @Szeto Basically: the harmonic series is divergent. We can choose signs such that the series converges to 0. $\endgroup$ – Parcly Taxel Feb 28 at 9:21
  • $\begingroup$ @ParclyTaxel I think you have to elaborate. $\endgroup$ – Szeto Feb 28 at 9:24
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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 $1,2,3,4,6,8,10,12,\dots$

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