找到一篇論文...
On a sum involving powers of reciprocals of
an arithmetical progression
http://www.kurims.kyoto-u.ac.jp/EMIS/journals/AMI/2007/ami2007-belbachir.pdf
他其實說下面這個題目的每一項都分別給一個次方，那加起來還是不會是整數。
用的方法其實就是傳統證1+1/2+...+1/n非整數的方法
只是他現在有一個很厲害的 Theorem 1.4
這我又再去查一下就資訊爆炸了...
※ 引述《darkseer (進入無限期公假)》之銘言：
: This problem is from AoPS.
: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26137
: n, a, and b are natural numbers. Prove that:
: 1 1 1
: --- + --- + ... + ---- isn't natural.
: a a+b a+nb
: By being natural, we mean that number is a positive integer.
: Your English may not be as poor as mine, but it takes me 5 mins to
: understand it orz...
--
r=e^theta
即使有改變，我始終如一。
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