Thread: Problem 48
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Old June 6th, 2008, 02:14 PM
Jacobsen Jacobsen is offline
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[FONT='Cambria Math','serif']My first thought was to try an inductive argument, but I had a lot of difficulty getting it going. I don’t think what I came up with is sound, but nevertheless I decided to post what I came up with.

Proof. It suffices to show that for all
n≥2; 1-1/2+1/3-…±1/n ∈ (0,1).
Let Pn denote the proposition that
1-1/2+1/3-…±1/(n-1) ∈ (0,1)
and
1-1/2+1/3-…±1/(n-1)±1/n∈ (0,1).
Then P3 is true since
1-1/2=1/2∈ (0,1)
and
1-1/2+1/3=5/6∈ (0,1)
Assume Pn is true and that n is even. Then
1-1/2+1/3-…+1/(n-1) ∈ (0,1)
and
1-1/2+1/3-…+1/(n-1)-1/n ∈ (0,1).
Because 1/(n+1) < 1/n, it follows from the inductive hypothesis that
1-1/2+1/3-…+1/(n-1)-1/n+1/(n+1) ∈ (0,1).
The case where n is odd is similar. So by the principle of mathematical induction, for all n ≥ 3, Pn is true and hence for all n ≥ 2, 1-1/2 +1/3 -…±1/n ∈ (0,1) and hence not an integer. //

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