送交者: lala 于 December 21, 2009 13:47:58:
回答: 今天上了一天班，从早十点上到晚八点半，还多亏有lala帮忙，否则。。。。。。 由 老六 于 December 20, 2009 22:43:55:
(1!+2!+...+N!)/(10^2+10^4+....+10^(2N))
> N!/(100/99)10^(2N)
> 0.99 ((2piN)^0.5) ((N/100e)^N) (by Stirling's formula)
Clearly, (1!+2!+...+N!)/(10^2+10^4+....+10^(2N)) > 1 whenever N > 100e
(i.e., N > 271).
You now can use calculator to calculate
0.99 ((2X3.14159X271)^0.5) ((271/100e)^271) = 17.86828381 > 1
0.99 ((2X3.14159X270)^0.5) ((270/100e)^270) = 6.593455162 > 1
0.99 ((2X3.14159X269)^0.5) ((269/100e)^269) = 2.442017627 > 1
0.99 ((2X3.14159X268)^0.5) ((268/100e)^268) = 0.907812193 < 1
(1!+2!+...+ 268!)/(10^2+10^4+....+10^(2X268))
< (1 + (2/268))268!/10^(2X268)
< (1 + (2/268))((2X3.1416X268)^0.5) (e^(1/(12X268))) ((268/100e)^268)
(by Stirling's formula again)
= 1.007462687 X 41.03532137 X 1.000310994 X 0.022346199
= 0.924113937 < 1.