[FONT='Cambria Math','serif']My first thought was to try an inductive argument, but I had a lot of difficulty getting it going. I dont 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
Let Pn denote the proposition that
±1/n ∈ (0,1).
±1/(n-1) ∈ (0,1)
Then P3 is true since
Assume Pn is true and that n is even. Then
+1/(n-1) ∈ (0,1)
Because 1/(n+1) < 1/n, it follows from the inductive hypothesis that
+1/(n-1)-1/n ∈ (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. //
+1/(n-1)-1/n+1/(n+1) ∈ (0,1).