- #1

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## Main Question or Discussion Point

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

[tex]

a_1 = 1,

a_2 = \int^2_1 \frac{dx}{x},

a_3 = \frac{1}{2},

a_4 = \int^3_2 \frac{dx}{x} ,

...

[/tex]

When generalized gives For any natural number n,

[tex]

a_{2n-1} = \frac{1}{n}

[/tex]

[tex]

a_{2n} = \int^{n+1}_n \frac{dx}{x} = \ln x |^{n+1}_{n} = \ln \frac{n+1}{n}

[/tex]

is decreasing, that is,

[tex]

\frac{1}{n} > \ln \frac{n+1}{n} > \frac{1}{n+1}

[/tex]

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

[tex]

a_1 = 1,

a_2 = \int^2_1 \frac{dx}{x},

a_3 = \frac{1}{2},

a_4 = \int^3_2 \frac{dx}{x} ,

...

[/tex]

When generalized gives For any natural number n,

[tex]

a_{2n-1} = \frac{1}{n}

[/tex]

[tex]

a_{2n} = \int^{n+1}_n \frac{dx}{x} = \ln x |^{n+1}_{n} = \ln \frac{n+1}{n}

[/tex]

is decreasing, that is,

[tex]

\frac{1}{n} > \ln \frac{n+1}{n} > \frac{1}{n+1}

[/tex]

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