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I encountered this problem in one of my math lecture notebooks; Our professor didnt show how it was done, so that leaves me clueless. The problem was to show that the sequence {a_n} defined by
<br /> a_1 = 1, <br /> a_2 = \int^2_1 \frac{dx}{x},<br /> a_3 = \frac{1}{2},<br /> a_4 = \int^3_2 \frac{dx}{x} ,<br /> ...<br />
When generalized gives For any natural number n,
<br /> a_{2n-1} = \frac{1}{n} <br />
<br /> a_{2n} = \int^{n+1}_n \frac{dx}{x} = \ln x |^{n+1}_{n} = \ln \frac{n+1}{n}<br />
is decreasing, that is,
<br /> \frac{1}{n} > \ln \frac{n+1}{n} > \frac{1}{n+1}<br />
I've tried math induction but I'm stuck at the (ii) part of math induction, and i tried comparing their derivatives, but I can't conclude anything from doing so. I've tried to compute for their areas, but that got me nowhere. I've graphed their functions using a graphing program, and I saw that it is true, but I would like know how i can prove this without graphing...
thanx in advance for all help and advice on my problem
<br /> a_1 = 1, <br /> a_2 = \int^2_1 \frac{dx}{x},<br /> a_3 = \frac{1}{2},<br /> a_4 = \int^3_2 \frac{dx}{x} ,<br /> ...<br />
When generalized gives For any natural number n,
<br /> a_{2n-1} = \frac{1}{n} <br />
<br /> a_{2n} = \int^{n+1}_n \frac{dx}{x} = \ln x |^{n+1}_{n} = \ln \frac{n+1}{n}<br />
is decreasing, that is,
<br /> \frac{1}{n} > \ln \frac{n+1}{n} > \frac{1}{n+1}<br />
I've tried math induction but I'm stuck at the (ii) part of math induction, and i tried comparing their derivatives, but I can't conclude anything from doing so. I've tried to compute for their areas, but that got me nowhere. I've graphed their functions using a graphing program, and I saw that it is true, but I would like know how i can prove this without graphing...
thanx in advance for all help and advice on my problem
