5. Multiple Choice 1 point Use the Limit Comparison Test to determine the convergence or divergence of the series: $$ \sum_{n=2}^{\infty} \frac{1}{\sqrt{n^{2}-1}} $$

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1. **Identify a Comparable Series:**

For large $$ n $$, note that
   $$
   \sqrt{n^2 - 1} \sim \sqrt{n^2} = n.
   $$
   Thus, we compare with the series
   $$
   \sum_{n=2}^{\infty} \frac{1}{n},
   $$
   which is a $$ p $$-series with $$ p=1 $$ and is known to diverge (harmonic series).

2. **Compute the Limit for the Limit Comparison Test:**

Let
   $$
   a_n=\frac{1}{\sqrt{n^2-1}} \quad \text{and} \quad b_n=\frac{1}{n}.
   $$
   We compute
   $$
   L = \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{\frac{1}{\sqrt{n^2-1}}}{\frac{1}{n}} = \lim_{n\to\infty}\frac{n}{\sqrt{n^2-1}}.
   $$

3. **Simplify the Limit:**

Divide the numerator and the denominator inside the square root by $$ n $$:
   $$
   \frac{n}{\sqrt{n^2-1}} = \frac{n}{n\sqrt{1-\frac{1}{n^2}}} = \frac{1}{\sqrt{1-\frac{1}{n^2}}}.
   $$
   Taking the limit as $$ n \to \infty $$,
   $$
   L = \frac{1}{\sqrt{1-0}} = 1.
   $$

4. **Conclude by the Limit Comparison Test:**

Since $$ L=1 $$ (a finite positive number) and the comparison series $$ \sum_{n=2}^{\infty} \frac{1}{n} $$ diverges, by the Limit Comparison Test, the series
   $$
   \sum_{n=2}^{\infty}\frac{1}{\sqrt{n^2-1}}
   $$
   also diverges.
The series $$ \sum_{n=2}^{\infty} \frac{1}{\sqrt{n^{2}-1}} $$ diverges.

Multiple Choice 1 point
Use the Limit Comparison Test to determine the convergence or divergence of the series:

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