Suppose delta > 0
Prove that there exists a positive integer n such that
0 < 1/(4n+1)pi/2 < delta
and sin( (4n+1)pi)/2) = 1
How do you do this?
It is very hard to read you post. I assume that it is $\displaystyle \frac{1}{\frac{(4n+1)\pi}{2}}$
Do you understand that the sequence $\displaystyle \left(\frac{2}{(4n+1)\pi}\right)\to 0$?
Then if $\displaystyle \delta >0$ that almost all of the terms of that sequence are less than $\displaystyle \delta$.