簡單有關e數學問題 (20點)

e (x=1)

e = 1 + 1 + 1/2! + 1/3! + 1/4! + ....
s(n) = 1 + 1 + 1/2! + 1/3! + 1/4! + .... + 1/(n-1)!

Compare e and s(n). The term of s(n) is less than e, so s(n) < e (x=1)

s*(n)
= s(n)＋1/(n-1)!
= 1 + 1 + 1/2! + 1/3! + 1/4! + .... + 1/(n-1)! + 1/(n-1)!

Consider
s*(n) - e
=1/(n-1)!-(1/n!+1/(n+1)!+...)
=1/(n-1)!-(1/n!)(1+1/(n+1)+1/(n+1)(n+2)...)
=(1/n!)[n-(1+1/(n+1)+1/(n+1)(n+2)...)]
＞(1/n!)[n-(1+1/(n+1)+1/(n+1)(n+1)...)]
=(1/n!)[n-(n+1)/n]
=(1/n!)[n-1+1/n]
＞0

s*(n) > e

elementary analysis....?
then, infinite series 的 convergence 要用 limit 表示

2008-06-22 15:23:31 補充：
如果係 A-Level Pure Maths 題目，好似myisland8132 ( 知識長 )的答案就得。
如果係數學專科，就要嚴謹一點，要用
e = lim{ N approaches +infinity } SUM{r from 0 to N} 1/r!
比較 finite 時的情況, 再take limit。
.
因為加法係 binary operation, 係無得將無限個數加埋的...