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 $L$ .

Stumbling Robot

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

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