Does the sequence P_n converge to (1,0)?

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SUMMARY

The sequence P_n = [(n+1)/n, [(-1)^n]/n] converges to the point p = (1, 0). This conclusion is reached by demonstrating that for any neighborhood U around p with radius r > 0, there exists an index N such that for all n ≥ N, the terms of the sequence P_n fall within U. The proof involves showing that the expression |P_n - p| can be made less than r, specifically by manipulating the inequality involving the square root of the sum of squares of the components of P_n and p.

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Homework Statement



Show that the sequence P_n = [(n+1)/n, [(-1)^n]/n] converges.

Homework Equations



A sequence p_n converges to a point p if and only if every neighborhood about p contains all the terms p_n for sufficiently large indices n; to any neighborhood U about p, there corresponds an index N such that p_n exists in U whenever n > or = N.

The Attempt at a Solution



The sequence converges to the point p = (1 0). Let U be the open ball around p with radius r>0. Need to show that |p_n - p|< r for all n > or = N i.e. ...

|((n+1)/n, ((-1)^n)/n) - (1, 0)| < r

sqrt of ((n+1)/n - 1)^2 + (((-1)^n)/n)^2 < r

Now I assume the n and N come into play, but I don't know how.
 
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Simply further. You should get something like 2/n^2 under the square root.
 

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