Determine the convergence or divergence of f(n) = n(-1)n

Consider the function $f(n)$ defined by

$f(n) = n^{(-1)^n}.$

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

The sequence $\{ f_n \}$ diverges.
Proof. Suppose otherwise, that there exists a real number $L$ and a positive integer $N$ such that

$|f(n) - L| < \varepsilon \qquad \text{for all } \varepsilon > 0, \quad \text{for all } n > N.$

Since $N$ is positive, we know $2N > N$ and $2N+1 > N$ . So,

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

Then,

$\begin{align*} && |2N - L| + \left| L - \frac{1}{2N+1} \right| &< 1 \\[9pt] \implies && \left| 2N - \frac{1}{2N+1} \right| &< 1 \\[9pt] \implies && 2N - \frac{1}{2N+1} &< 1 &(\text{since } 2N - \frac{1}{2N+1} > 0) \\[9pt] \implies && 2N - \frac{1}{2N} &< 1 \\[9pt] \implies && 2N - 1 &< 1 \\[9pt] \implies && 2N & < 2 \\[9pt] \implies && N &< 1. \end{align*}$

This contradicts that $N$ is a positive integer (since there are no positive integers less than 1). Hence, $\{ f_n \}$ must diverge $. \qquad \blacksquare$

Stumbling Robot

Determine the convergence or divergence of f(n) = n^(-1)ⁿ

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