What is $\frac{(-1)^n}{n}$ as $n$ approaches infinity? What is the limit of $\frac{(-1)^n}{ n}$ as $n$ approaches positive infinity? I can see how it would converge to zero, as the denominator swiftly over powers the numerator. However, the top goes into the imaginary plane for non-integer $n$. Furthermore, since the limit as $x$ goes toward infinity of $\sin(x)$ is DNE, would the same logic apply here? Is the answer $0$ or DNE?

$$ \left|\frac{(-1)^n}{n}\right|\leq \frac{1}{n} \longrightarrow 0 $$