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a) {2^n/n!}

b) {(n+3)/(6n-1)}

c) {4^(1/n)}

2007-04-04 04:58:43 · 1 answers · asked by Anonymous in Science & Mathematics Mathematics

1 answers

taking a_n+1/a_n = 2/(n+1)<1 so it is monotone decreasing
b) (n+4)/(6n+5) -(n+3)/(6n-1)=
=[(n+4)(6n-1)-(6n+5)(n+3)]/(6n+5)(6n-1)=
(6n^2-n+24n-4-6n^2-18n-5n-15)/36n^2+24n-5
=-19/(n^2+24n-5) <0 so it is monotone decreasing

c) 4^1(n+1) ? 4^1/n
Taking log base 4
1/(n+1)<1/n so 4^1/(n+1)< 4^1/n as the base is >1. Monotone decreasing

2007-04-04 05:25:37 · answer #1 · answered by santmann2002 7 · 0 0

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