1,662
edits
Citation bot (talk  contribs) (Add: bibcode, arxiv. Removed URL that duplicated unique identifier.  You can use this bot yourself. Report bugs here.  Activated by Headbomb  via #UCB_webform) 
(→As an infinite series: minor simplification: 3  \sum_{k=2}^\infty \frac{1}{k! (k1) k}) 

Consideration of how to put upper bounds on ''e'' leads to this descending series:
:<math>e = 3
which gives at least one correct (or rounded up) digit per term. That is, if 1 ≤ ''n'', then
:<math>e < 3
More generally, if ''x'' is not in {2, 3, 4, 5, ...}, then
:<math>e^x = \frac{2+x}{2x} + \sum_{k=2}^\infty \frac{ x^{k+1}}{k! (kx) (k+1x)} \,.</math>
