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

[solakis1]

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:

$$|2-(-1)^n-l|\geq a$$

 

[HallsofIvy]

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.

 

[solakis1]

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 ??