MHB High school inequality |2−(−1)n−l|≥a

AI Thread Summary
The discussion focuses on proving the existence of a positive constant \( a \) such that for any natural number \( k \), there exists an integer \( n \) greater than or equal to \( k \) satisfying the inequality \( |2 - (-1)^n - l| \geq a \). Participants clarify that for any integer \( n \), the expression \( (-1)^n \) results in either 1 or -1, leading to two cases for the absolute value: \( |1 - l| \) or \( |3 - l| \). The consensus is that both odd and even integers exist beyond any given \( k \), supporting the claim. The discussion emphasizes the need for a clearer understanding of the inequality's implications. Overall, the mathematical premise is deemed valid with proper interpretation.
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Given any real No $$l$$,then prove,that there exist $$a>0$$ such that ,for all natural Nos $$k$$ there exist $$n\geq k$$
such that:

$$|2-(-1)^n-l|\geq a$$
 
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Frankly, that doesn't make much sense. For any integer, n, (-1)^n is either 1 (n even) or -1 (n odd). So |2- (-1)^n- l| is either |2- 1- l|= |1- l| or |2+ 1- l|= |3- l|. Of course there exist both odd and even numbers larger than any given k.
 
HallsofIvy said:
Frankly, that doesn't make much sense. For any integer, n, (-1)^n is either 1 (n even) or -1 (n odd). So |2- (-1)^n- l| is either |2- 1- l|= |1- l| or |2+ 1- l|= |3- l|. Of course there exist both odd and even numbers larger than any given k.

Given any $$l$$ the OP is looking for $$a>0$$,such that:

$$\forall k(k\in N\Longrightarrow\exists n(n\geq k\wedge |2-(-1)^n-l|\geq a))$$

Does it make sense now ??
 
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