Determine the convergence or divergence of f(n) = (-1)n / n + (1+(-1)n) / 2

Consider the function $f(n)$ defined by

$f(n) = \frac{(-1)^n}{n} + \frac{1+(-1)^n}{2}.$

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 a positive integer we have $2N > N$ and $2N+1>N$ . Thus,

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

Therefore, using the triangle inequality we have

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

But, this is false for any $0 < \varepsilon < \frac{1}{2}$ . Hence, there can be no such $L$ , so $\{ f(n) \}$ diverges $. \qquad \blacksquare$

Stumbling Robot

Determine the convergence or divergence of f(n) = (-1)ⁿ / n + (1+(-1)ⁿ) / 2

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