Re: [微積] 台大數研90高微考古題
※ 引述《tasukuchiyan (Tasuku)》之銘言：
: 4.(b)
: Let a(n)>0, a(n+1)/a(n)<=(1-2/n) for n>=3.
: Show that the series of a(n) is convergent.
consider
lim_{x→∞} [ln x - ln(x-2)]/[ln (x+1) - ln x]
= lim_{x→∞} [1/x - 1/(x-2)] / [1/(x+1) - 1/x]
= lim_{x→∞} (x-2-x)/[x(x-2)] / {(x-x-1)/[(x+1)x]}
= 2
hence, there exists N such that
[ln n - ln (n-2)] > (1.5) [ ln (n+1) - ln n] whenever n > N
n/(n-2) > [(n+1)/n]^(1.5)
1-2/n = (n-2)/n < [n/(n+1)]^{1.5}
that is,
a(n+1)/a(n) < [n/(n+1)]^{1.5}
hence,
a(N+2) < a(N+1) [(N+1)/(N+2)]^{1.5}
a(N+3) < a(N+2) [(N+2)/(N+3)]^{1.5} < a(N+1) [(N+1)/(N+3)]^1.5
.....
then
a(N+2) + a(N+3) + ....
< a(N+1) (N+1)^{1.5} [ 1/(N+2)^{1.5} + 1/(N+3)^{1.5} + 1/(N+4)^{1.5} + ...]
Convergent,
Done.
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