Series

BY AM>GM(a1不等於a2)

[1+n^-1)+(1+n^-1)+....+(1+n^-1)+1]/(1+n)>(1+n^-1)^[n/(n+1)]

(2+n)/(1+n)>(1+n^-1)^[n/(n+1)]

1+(1+n)^-1>(1+n^-1)^[n/(n+1)]

(1+n-1)n<[1+(1+n)-1]n+1-------------ans

2009-10-09 10:11:35 補充：
純不等式證明

2009-10-09 10:13:22 補充：
上面個個兄弟做得較煩喎@W@"

2009-10-09 10:13:24 補充：
上面個個兄弟做得較煩喎@W@"

001, You can have a more concise presentation by using summation signs. That will be more neat and easy to compare.

Good Job!!!

(1 + n-1)n = 1 + n(n-1) + nC2 n-2 + nC3 n-3 + ... + nCn n-n
= 1 + 1 + (n - 1)/(2! n) + (n - 1)(n - 2)/(3! n2) + ... + (n - 1)(n - 2)...(1)/(n! nn-1)
= 2 + [(n - 1)/n](1/2!) + [(n - 1)/n][(n - 2)/n](1/3!) + ... + [(n - 1)/n][(n - 2)/n] ... (1/n)(1/n!)
[1 + (1 + n)-1]n+1 = 1 + (n + 1)(n + 1)-1 + (n + 1)C2 (n + 1)-2 + (n + 1)C3 (n + 1)-3 + ... + (n + 1)Cn (n + 1)-n + (n + 1)C(n + 1) (n + 1)-n-1
= 1 + 1 + n/[2! (n + 1)] + n(n - 1)/[3! (n + 1)2] + ... + n(n - 1)...(2)/[n! (n + 1)n-1] + n(n - 1)...(1)/[(n + 1)! (n + 1)n]
= 2 + [n/(n + 1)](1/2!) + [n/(n + 1)][(n - 1)/(n + 1)](1/3!) + ... + [n/(n + 1)][(n - 1)/(n + 1)] ... [2/(n + 1)](1/n!) + [n/(n + 1)][(n - 1)/(n + 1)] ... [1/(n + 1)][1/(n + 1)!]
With:
(n - 1)/n < n/(n + 1) (since 1 - 1/n < 1 - 1/(n + 1))
(n - 2)/n < (n - 1)/(n + 1)
.
.
1/n < 2/(n + 1)
It concludes that (1 + n-1)n < [1 + (1 + n)-1]n+1.

2009-10-09 10:20:08 補充：
樓下, 你不能 assume 發問者 AM >= GM, 因為此題未必是 Pure Math 題目