Determine whether the sequence converges or diverges. If it converges, find the limit. (If an DNE.)
$$
\begin{array}{c}
a_{n}=\frac{(-1)^{n}}{8 \sqrt{n}} \
\lim _{n \rightarrow \infty} a_{n}=\square
\end{array}
$$
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∻Solution by Steps∻ ‖ step 1 ⋮ We start by considering the given sequence: $a_n = \frac{(-1)^n}{8 \sqrt{n}}$. ‖ ‖ step 2 ⋮ To determine if the sequence converges, we need to find the limit as $n$ approaches infinity: $\lim_{n o \infty} \frac{(-1)^n}{8 \sqrt{n}}$. ‖ ‖ step 3 ⋮ Observe that $(-1)^n$ oscillates between -1 and 1, but $8 \sqrt{n}$ grows without bound as $n$ approaches infinity. ‖ ‖ step 4 ⋮ Therefore, the numerator $(-1)^n$ is bounded, while the denominator $8 \sqrt{n}$ increases indefinitely. ‖ ‖ step 5 ⋮ As $n$ approaches infinity, the fraction $\frac{(-1)^n}{8 \sqrt{n}}$ approaches 0 because the denominator grows much faster than the numerator. ‖ ‖ step 6 ⋮ Thus, $\lim_{n o \infty} \frac{(-1)^n}{8 \sqrt{n}} = 0$. ‖ ∻Answer∻ ⚹ The sequence converges to 0. ⚹ ∻Key Concept∻ ⚹Convergence of a sequence⚹ ∻Explanation∻ ⚹A sequence converges if its terms approach a specific value as $n$ approaches infinity. In this case, the sequence $\frac{(-1)^n}{8 \sqrt{n}}$ converges to 0 because the denominator grows without bound while the numerator remains bounded.⚹◊What is the general formula for determining whether a sequence converges or diverges in mathematics?⍭ Generate me a similar question◊

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