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# e is irrational

We have the series

$e^{{-1}}=\sum_{{k=0}}^{\infty}{(-1)^{k}\over k!}$ |

Note that this is an alternating series and that the magnitudes of the terms decrease. Hence, for every integer $n>0$, we have the bound

$0<\left|\sum_{{k=0}}^{{n}}{(-1)^{k}\over k!}-e^{{-1}}\right|<{1\over(n+1)!},$ |

by the Leibniz’ estimate for alternating series. Assume that $e=n/m$, where $m$ and $n$ are integers and $n>0$. Then we would have

$0<\left|\sum_{{k=0}}^{{n}}{(-1)^{k}\over k!}-{m\over n}\right|<{1\over(n+1)!}.$ |

Multiplying both sides by $n!$, this would imply

$0<\left|\sum_{{k=0}}^{{n}}{(-1)^{k}n!\over k!}-m(n-1)!\right|<{1\over n+1},$ |

which is a contradiction because every term in the sum is an integer, but there are no integers between $0$ and $1/(n+1)$.

Related:

LeibnizEstimateForAlternatingSeries

Type of Math Object:

Proof

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Reference

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## Mathematics Subject Classification

11J82*no label found*11J72

*no label found*

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