Consider the function 
defined by
![\[ f(n) = \frac{1}{n} e^{-\frac{\pi i n}{2}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-326ca960b66d3bbc684fc7f5cfdda999_l3.png "Rendered by QuickLaTeX.com")
Determine the convergence or divergence of the sequence 
and if it converges determine its limit.
This sequence converges since
![\begin{align*} \lim_{n \to \infty} |f(n)| &= \lim_{n \to \infty} \left| \frac{1}{n} \cdot e^{-\frac{\pi i n}{2}} \right| \\[9pt] &= \lim_{n \to \infty} \left( \frac{1}{n} \cdot \left| e^{-\frac{\pi i n}{2}} \right| \right). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-921f021699bde768ad8488d4a70a04e8_l3.png "Rendered by QuickLaTeX.com")
But each of the limits in the product exists since
![\[ \lim_{n \to \infty} \frac{1}{n} = 0 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e7bbd2f135382d6e84973cbaae533ea2_l3.png "Rendered by QuickLaTeX.com")
and
![\[ \lim_{n \to \infty} \left| e^{-\frac{\pi i n}{2}} \right| = \lim_{n \to \infty} 1 = 1. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-cdf0fd3d786fe3a0734f4fb3e10e7639_l3.png "Rendered by QuickLaTeX.com")
Hence,
![\[ \lim_{n \to \infty} |f(n)| = 0 \cdot 1 = 0. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8508c1dbe12361215c5e09938b9bec04_l3.png "Rendered by QuickLaTeX.com")
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.