→ njru81l :實數的部分可以夾，用 [n]<=n<[n]+1 07/27 16:29
小弟只會證明n是正整數時，極限存在~Orz
遞增:
(1+1/n)^n
=1+C(n,1)*(1/n)+C(n,2)*(1/n)^2+...+C(n,n)*(1/n)^n
=1+1+[(n-1)/n]*[1/(2!)]+[(n-1)*(n-2)/(n*n)]*[1/(3!)]
+...+[(n-1)*(n-2)*...*1/(n*n*...*n)]*[1/(n!)]
<=1+1+[(n)/(n+1)]*[1/(2!)]+[(n)*(n-1)/((n+1)*(n+1))]*[1/(3!)]
+...+[n*(n-1)*...*2/((n+1)*(n+1)*...*(n+1))]*[1/(n!)]
<=1+1+[(n)/(n+1)]*[1/(2!)]+[(n)*(n-1)/((n+1)*(n+1))]*[1/(3!)]
+...+[n*(n-1)*...*2/((n+1)*(n+1)*...*(n+1))]*[1/(n!)]
+C(n+1,n+1)*[1/(n+1)]^(n+1)
=1+C(n+1,1)*[1/(n+1)]+C(n+1,2)*[1/(n+1)]^2+C(n+1,3)*[1/(n+1)]^3
+...+C(n+1,n)*[1/(n+1)]^n
+C(n+1,n+1)*[1/(n+1)]^(n+1)
=[1+1/(n+1)]^(n+1)
ps: (n-k)/n <= (n-k+1)/(n+1) for n,k are 正整數
有上界:
(1+1/n)^n
=1+C(n,1)*(1/n)+C(n,2)*(1/n)^2+...+C(n,n)*(1/n)^n
<=1+1/(1!)+1/(2!)+1/(3!)+......
<=1+1+1/2+1/(2^2)+......=3
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