It's actually 9\* (1 + 8 \* (1 + 7 \* (1 + 6 ..... + 1))) | Hacker News

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		tdinkar on Jan 24, 2014 \| parent \| context \| favorite \| on: Functional Programming 101 with Haskell It's actually 9* (1 + 8 * (1 + 7 * (1 + 6 ..... + 1)))

winstonsmith on Jan 24, 2014 [–]

= [9!e] - 1

(because your expression expands to 9 + 9(8) + 9(8)(7) + ... 9! = 9!(1+1/1 + 1/2! + ... 1/8!) = -1 + 9! e - (1/10 + 1/(10 11) + ...))

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