Use the limit comparison test to determine if $$ \sum_{n=2}^{\infty} \frac{2}{4 n^{\frac{1}{2}}-4 n^{\frac{1}{6}}-1} $$ converges or diverges, and justify your answer. Answer Attempt 1 out of 2 Apply the comparison test with the series $$ \sum_{n=2}^{\infty} \frac{1}{n^{p}} $$ where $$ p= $$ $$ \frac{1}{2} $$. If $$ a_{n}=\frac{2}{4 n^{\frac{1}{2}}-4 n^{\frac{1}{6}}-1} $$ and $$ b_{n}=\frac{1}{n^{\frac{1}{2}}} $$, then $$ \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\square $$. Since $$ a_{n}, b_{n}0 $$ and the limit is a finite and positive (non-zero) number, the limit comparison test applies. $$ \sum_{n=2}^{\infty} \frac{1}{n^{\frac{1}{2}}} $$ diverges since a p-series will diverge if and only if $$ \mathrm{p}

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Use the limit comparison test to determine if $$ \sum_{n=2}^{\infty} \frac{2}{4 n^{\frac{1}{2}}-4 n^{\frac{1}{6}}-1} $$ converges or diverges, and justify your answer. Answer Attempt 1 out of 2 Apply the comparison test with the series $$ \sum_{n=2}^{\infty} \frac{1}{n^{p}} $$ where $$ p= $$ $$ \frac{1}{2} $$. If $$ a_{n}=\frac{2}{4 n^{\frac{1}{2}}-4 n^{\frac{1}{6}}-1} $$ and $$ b_{n}=\frac{1}{n^{\frac{1}{2}}} $$, then $$ \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\square $$. Since $$ a_{n}, b_{n}>0 $$ and the limit is a finite and positive (non-zero) number, the limit comparison test applies. $$ \sum_{n=2}^{\infty} \frac{1}{n^{\frac{1}{2}}} $$ diverges since a p-series will diverge if and only if $$ \mathrm{p}<1 \vee $$. Therefore, $$ \sum_{n=2}^{\infty} \frac{2}{4 n^{\frac{1}{2}}-4 n^{\frac{1}{6}}-1} $$ diverges

UpStudy AI · Accepted Answer

We are given
$$
a_n = \frac{2}{4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1}
$$
and we compare it with the series
$$
b_n = \frac{1}{n^{\frac{1}{2}}}.
$$

1. **Determine the asymptotic behavior:**

As $$ n \to \infty $$, the dominant term in the denominator of $$ a_n $$ is $$ 4n^{\frac{1}{2}} $$, so we have
   $$
   4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1 \sim 4n^{\frac{1}{2}},
   $$
   which suggests
   $$
   a_n \sim \frac{2}{4n^{\frac{1}{2}}} = \frac{1}{2} \cdot \frac{1}{n^{\frac{1}{2}}}.
   $$

2. **Compute the limit for the limit comparison test:**

We calculate
   $$
   L = \lim_{n \rightarrow \infty} \frac{a_n}{b_n} 
   = \lim_{n \rightarrow \infty} \frac{\frac{2}{4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1}}{\frac{1}{n^{\frac{1}{2}}}}
   = \lim_{n \rightarrow \infty} \frac{2n^{\frac{1}{2}}}{4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1}.
   $$

Factor out $$ n^{\frac{1}{2}} $$ from the denominator:
   $$
   4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1 = n^{\frac{1}{2}} \left( 4 - \frac{4}{n^{\frac{1}{3}}} - \frac{1}{n^{\frac{1}{2}}} \right).
   $$

Then the limit becomes:
   $$
   L = \lim_{n \rightarrow \infty} \frac{2n^{\frac{1}{2}}}{n^{\frac{1}{2}} \left( 4 - \frac{4}{n^{\frac{1}{3}}} - \frac{1}{n^{\frac{1}{2}}} \right)}
   = \lim_{n \rightarrow \infty} \frac{2}{4 - \frac{4}{n^{\frac{1}{3}}} - \frac{1}{n^{\frac{1}{2}}}}
   = \frac{2}{4} = \frac{1}{2}.
   $$

3. **Apply the limit comparison test:**

Since
   $$
   0 < L = \frac{1}{2} < \infty,
   $$
   the limit comparison test tells us that the series
   $$
   \sum_{n=2}^{\infty} a_n
   $$
   and
   $$
   \sum_{n=2}^{\infty} b_n
   $$
   either both converge or both diverge.

4. **Determine the behavior of the comparison series:**

The series
   $$
   \sum_{n=2}^{\infty} \frac{1}{n^{\frac{1}{2}}}
   $$
   is a $$ p $$-series with $$ p = \frac{1}{2} $$. Since $$ p \leq 1 $$, this series diverges.

5. **Conclusion:**

By the limit comparison test, the original series
   $$
   \sum_{n=2}^{\infty} \frac{2}{4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1}
   $$
   also diverges.
The series $$ \sum_{n=2}^{\infty} \frac{2}{4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1} $$ diverges because, using the limit comparison test with $$ \sum_{n=2}^{\infty} \frac{1}{n^{\frac{1}{2}}} $$, both series behave similarly as $$ n $$ approaches infinity, and since the comparison series diverges, the original series also diverges.

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