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

by RoRi

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.

One comment

  1. William C says:

    It mightv’e been simpler to look at the magnitudes of n and e^(-piin/2). Since the first is constant at one and the other approaches infinity, the overall magnitude approaches infinity, implying the limit approaches infinity and the sequence diverges (someone let me know if this doesn’t work for some reason)

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