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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?

pjs36
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Script Kitty
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2 Answers2

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

Hamza
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What matters, even if we consider real $n$ instead of integer $n$, is that $|(-1)^n| =|e^{\pi i n}| =|\cos(\pi n)+i\sin(\pi n)| =1 $.

Therefore, as Hamza wrote, $|\frac{(-1)^n}{n}| =|\frac{1}{n}| \to 0 $ as $n \to \infty$ for real or integer $n$.

marty cohen
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