送交者: lala 于 December 21, 2009 22:48:14:
回答: Finally, 老六 got a problem worth of lala's attention! :)) 由 lakeblue 于 December 21, 2009 20:24:27:
We show that
(*) 1! + 2! + ... + N! < (1 + (2/N))N!, N ≥ 1.
In fact, for all N ≥ 1,
1! + 2! + ... + N!
= N!((1!/N!) + (2!/N!) + ... + ((N-2)!/N!) + ((N-1)!/N!) + 1)
≤ N!((1/((N-1)N)) + (1/((N-1)N)) + ... + (1/((N-1)N)) + (1/N) + 1)
< N!(((N-1)/((N-1)N)) + (1/N) + 1)
= N!(1 + (2/N)),
i.e., (*) holds.