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Jun 6, 2016 at 16:17 comment added Klaus Draeger Note that the sum expands into $n!(\frac{1}{0!}+\dots+\frac{1}{n!})$; the sum in parentheses converges to $e$ from below, and can easily be shown to be in $(e-\frac{1}{n!},e)$.
Jun 6, 2016 at 14:09 comment added Arun Sen Jeffrey - Thank you so much for the answer. Btw, do you have a reference for the proof?
Jun 6, 2016 at 12:13 history answered Jeffrey Shallit CC BY-SA 3.0