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

Consider the function f(n) defined by

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

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


The sequence \{ f(n) \} diverges.
Proof. First, we use DeMoivre’s theorem to write,

    \begin{align*}  f(n) &= e^{-\frac{\pi i n}{2}} \\[9pt]  &= \left( e^{-\frac{\pi i}{2}} \right)^n \\[9pt]  &= \cos \left( \frac{-n \pi}{2} \right) - i \sin \left( \frac{n \pi}{2} \right). \end{align*}

(Note: On page 380 Apostol claims (without proof) that the sequence defined by f(n) = \sin \left( \frac{n \pi}{2} \right) is divergent. If you want to accept that then this sequence will diverge since a complex-valued sequence f(n) = u(n) + i v(n) diverges if and only if the sequences defined by the functions u(n) and v(n) converge. Since it is a good exercise (and maybe Apostol wanted us to prove it ourselves) we can prove this is divergent from the definition.)

Suppose, for the sake of contradiction, that the sequence \{ f (n) \} converges to a limit L. Then we know for every \varepsilon > 0 there exists a positive integer N such that for all n > N we have

    \[ |f(n) - L| < \varepsilon. \]

Since N is positive we know 4N > N and 4N+2 > N. Hence, taking \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 |\cos (2N \pi) - i \sin (2N \pi) - L| &< \frac{1}{2} & \text{and} && |\cos (2N \pi + \pi) - i \sin (2N \pi + \pi) -L| &< \frac{1}{2} \\[9pt]  \implies |1 - L| &< \frac{1}{2} & \text{and} && |1+L| &< \frac{1}{2}. \end{align*}

Adding these two expressions and using the triangle inequality,

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

a contradiction. Hence, the sequence \{ f(n) \} diverges. \qquad \blacksquare

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