High school inequality |2−(−1)n−l|≥a

[solakis1]

Given any real No \(\displaystyle l\),then prove,that there exist \(\displaystyle a>0\) such that ,for all natural Nos \(\displaystyle k\) there exist \(\displaystyle n\geq k\)
such that:

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

 

[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 \(\displaystyle l\) the OP is looking for \(\displaystyle a>0\),such that:

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

Does it make sense now ??