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Determine the convergence or divergence of f(n) = ne-πin / 2

Consider the function f(n) defined by

    \[ f(n) = ne^{-\frac{\pi i n}{2}}. \]

Determine whether \{ f(n) \} converges or diverges, and if it converges find its limit.


This sequence diverges.
Proof. We saw in this exercise (Section 10.4, Exercise #20) that the sequence defined by f(n) = e^{-\frac{\pi i n}{2}} diverges. We could use this to show that this sequence diverges (since for n = 4k we have f(n) = n and so \{ f(n) \} cannot approach a finite limit). For practice, we can also prove this directly from the definition by contradiction as follows. Suppose there exists a real number L and a positive integer N such that for all n > N and all \varepsilon > 0 we have

    \[ |f(n) - L| < \varepsilon. \]

Since N is positive we know 4N > N and 4N+2 > N. So, letting \varepsilon = \frac{1}{2}, we have

    \begin{align*}   && | f(4N) - L | &< \frac{1}{2} & \text{and} && |f(4N+2) - L| &< \frac{1}{2} \\[9pt]  \implies && | 4N - L | &< \frac{1}{2} & \text{and} && |-(4N+2) - L| &< \frac{1}{2} \\[9pt]   && && \implies && |4N+2+L| &< \frac{1}{2}. \end{align*}

Adding these two expressions and using the triangle inequality we have,

    \begin{align*}  && |4N-L| + |4N+2+L| &< 1 \\[9pt]  \implies && |4N - L + 4N + 2 + L| &< 1 \\[9pt]  \implies && |8N + 2| &< 1 \\[9pt]  \implies && 2 &< 1. \end{align*}

This is a contradiction. Hence, the sequence \{ f(n) \} does not tend to a limit L.

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