Consider the function defined by

Determine the convergence or divergence of the sequence and if it converges determine its limit.

This sequence converges since

But each of the limits in the product exists since

and

Hence,

Therefore, (as we saw in this exercise, if is complex valued and then the sequence as well). Hence, the sequence converges to 0.

e^((-pi)in/2) diverges. It does not approach 1 as n approaches infinity.

You are right!

But the statement is about the absolute value (modulus) of $e^{-in\frac{pi}{2}}$ e.g. $|e^{-in\frac{pi}{2}}|$ not the complex number $e^{-in\frac{pi}{2}}$ itself.