[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

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

and

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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