We are tasked with using the Limit Comparison Test to determine whether the series

$\sum_{n=1}^{\infty} \frac{3n^{\frac{5}{2}} + 6}{2n^3 - 2n^2 + 5}$

converges or diverges.

Step 1: Identify the dominant terms

The first step in applying the Limit Comparison Test is to identify the dominant terms in the numerator and the denominator for large $n$.

• In the numerator, the dominant term is $3n^{\frac{5}{2}}$ because it grows faster than the constant $6$.
• In the denominator, the dominant term is $2n^3$ because it grows faster than $-2n^2$ and the constant $5$.

Thus, for large $n$, the series behaves similarly to:

$\frac{3n^{\frac{5}{2}}}{2n^3}$

Step 2: Simplify the comparison series

Now, we simplify this expression:

$\frac{3n^{\frac{5}{2}}}{2n^3} = \frac{3}{2} \cdot n^{\frac{5}{2} - 3} = \frac{3}{2} \cdot n^{-\frac{1}{2}}$

So, the series behaves like:

$\sum_{n=1}^{\infty} \frac{3}{2} \cdot n^{-\frac{1}{2}}$

Step 3: Choose a comparison series

Now we compare the given series with the p-series:

$\sum_{n=1}^{\infty} \frac{1}{n^p}$

Here, the exponent $p$ is $\frac{1}{2}$.

Step 4: Apply the Limit Comparison Test

The Limit Comparison Test states that if we compute the limit:

$L = \lim_{n \to \infty} \frac{a_n}{b_n}$

where $a_n$ is the original series' general term, and $b_n$ is the general term of the comparison series, and $L$ is a positive finite number, then both series either converge or diverge together.

We already know that:

• $a_n = \frac{3n^{\frac{5}{2}} + 6}{2n^3 - 2n^2 + 5}$ behaves like $\frac{3}{2} n^{-\frac{1}{2}}$ for large $n$.
• The comparison series $b_n = \frac{1}{n^{\frac{1}{2}}}$.

Now, we compute the limit:

$\lim_{n \to \infty} \frac{\frac{3n^{\frac{5}{2}} + 6}{2n^3 - 2n^2 + 5}}{\frac{1}{n^{\frac{1}{2}}}} = \lim_{n \to \infty} \frac{3n^{\frac{5}{2}} + 6}{2n^3 - 2n^2 + 5} \cdot n^{\frac{1}{2}}$

For large $n$, this simplifies to:

$\lim_{n \to \infty} \frac{3n^{\frac{5}{2}} \cdot n^{\frac{1}{2}}}{2n^3} = \lim_{n \to \infty} \frac{3n^3}{2n^3} = \frac{3}{2}$

Thus, the limit is $\frac{3}{2}$, which is a positive finite number.

Step 5: Conclusion using the Limit Comparison Test

Since the limit is positive and finite, and we are comparing with a p-series $\sum \frac{1}{n^{\frac{1}{2}}}$, we need to check the behavior of the p-series.

For a p-series $\sum \frac{1}{n^p}$ to converge, we need $p > 1$. In our case, $p = \frac{1}{2}$, which is less than 1. Therefore, the comparison series diverges.

Since the original series behaves similarly to this divergent p-series, the given series also diverges.

Final Answer:

The series $\sum_{n=1}^{\infty} \frac{3n^{\frac{5}{2}} + 6}{2n^3 - 2n^2 + 5}$ diverges.

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Would you like further details or clarifications on any step?

Here are some related questions for further exploration:

1. How does the Limit Comparison Test work in other series?
2. What is the p-series test, and how do we apply it?
3. What is the behavior of the series $\sum \frac{1}{n^p}$ for different values of $p$?
4. Can the Limit Comparison Test be used for all types of series?
5. How does the Ratio Test compare to the Limit Comparison Test in determining convergence?

Tip: Always focus on the dominant terms in both the numerator and the denominator when applying the Limit Comparison Test. This will simplify the process and give you a clearer understanding of the series' behavior.