極限e的序列
熟知:
(1+1/n)^n ↑e,(1+1/n)^(n+1)↓e (*)
事實上有
(1+1/n)^(n+1/2)↓e (重點是遞減)
證明:
考慮函數 f(x) = (x+1/2) log (1+ 1/x)
f'(x)
= log (1 +1/x) - (x+1/2)/[x(x+1)]
= log (1 +1/x) - 1/2 {1/x + 1/(x+1)} → 0 as x→無限大
f"(x)
= -1/[x(x+1)] + 1/2{1/x^2 + 1/(x+1)^2}
=1/2{1/x - 1/(x+1)}^2 > 0,故f'↑0,f'<=0,f↓
請問有沒有不用微積分的方法?
(例如(*)的兩個序列的增減都可用算幾不等式得到)
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r=e^theta
即使有改變,我始終如一。
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