The problem asks to determine the convergence or divergence of four series using the Limit Comparison Test. 1. $\sum_{n=1}^{\infty} \frac{n}{n^2 + 2n + 3}$

1. Problem Description

The problem asks to determine the convergence or divergence of four series using the Limit Comparison Test.

1. $\sum_{n=1}^{\infty} \frac{n}{n^2 + 2n + 3}$

2. $\sum_{n=1}^{\infty} \frac{3n + 1}{n^3 - 4}$

3. $\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n+1}}$

4. $\sum_{n=1}^{\infty} \frac{\sqrt{2n+1}}{n^2}$

2. Solution Steps

Problem 1: $\sum_{n=1}^{\infty} \frac{n}{n^2 + 2n + 3}$

We compare this series with $\sum_{n=1}^{\infty} \frac{1}{n}$, which is a divergent harmonic series. Let $a_n = \frac{n}{n^2 + 2n + 3}$ and $b_n = \frac{1}{n}$.

$\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{\frac{n}{n^2 + 2n + 3}}{\frac{1}{n}} = \lim_{n\to\infty} \frac{n^2}{n^2 + 2n + 3} = \lim_{n\to\infty} \frac{1}{1 + \frac{2}{n} + \frac{3}{n^2}} = 1$

Since the limit is a positive finite number (1), and $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, then $\sum_{n=1}^{\infty} \frac{n}{n^2 + 2n + 3}$ also diverges.

Problem 2: $\sum_{n=1}^{\infty} \frac{3n + 1}{n^3 - 4}$

We compare this series with $\sum_{n=1}^{\infty} \frac{1}{n^2}$, which is a convergent p-series with $p = 2 > 1$. Let $a_n = \frac{3n + 1}{n^3 - 4}$ and $b_n = \frac{1}{n^2}$.

$\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{\frac{3n + 1}{n^3 - 4}}{\frac{1}{n^2}} = \lim_{n\to\infty} \frac{3n^3 + n^2}{n^3 - 4} = \lim_{n\to\infty} \frac{3 + \frac{1}{n}}{1 - \frac{4}{n^3}} = 3$

Since the limit is a positive finite number (3), and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, then $\sum_{n=1}^{\infty} \frac{3n + 1}{n^3 - 4}$ also converges.

Problem 3: $\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n+1}}$

We compare this series with $\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$, which is a convergent p-series with $p = \frac{3}{2} > 1$. Let $a_n = \frac{1}{n\sqrt{n+1}}$ and $b_n = \frac{1}{n\sqrt{n}} = \frac{1}{n^{3/2}}$.

$\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{\frac{1}{n\sqrt{n+1}}}{\frac{1}{n\sqrt{n}}} = \lim_{n\to\infty} \frac{n\sqrt{n}}{n\sqrt{n+1}} = \lim_{n\to\infty} \frac{\sqrt{n}}{\sqrt{n+1}} = \lim_{n\to\infty} \sqrt{\frac{n}{n+1}} = \lim_{n\to\infty} \sqrt{\frac{1}{1 + \frac{1}{n}}} = \sqrt{1} = 1$

Since the limit is a positive finite number (1), and $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ converges, then $\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n+1}}$ also converges.

Problem 4: $\sum_{n=1}^{\infty} \frac{\sqrt{2n+1}}{n^2}$

We compare this series with $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2} = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$, which is a convergent p-series with $p = \frac{3}{2} > 1$. Let $a_n = \frac{\sqrt{2n+1}}{n^2}$ and $b_n = \frac{\sqrt{n}}{n^2} = \frac{1}{n^{3/2}}$.

$\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{\frac{\sqrt{2n+1}}{n^2}}{\frac{\sqrt{n}}{n^2}} = \lim_{n\to\infty} \frac{\sqrt{2n+1}}{\sqrt{n}} = \lim_{n\to\infty} \sqrt{\frac{2n+1}{n}} = \lim_{n\to\infty} \sqrt{2 + \frac{1}{n}} = \sqrt{2}$

Since the limit is a positive finite number ($\sqrt{2}$), and $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ converges, then $\sum_{n=1}^{\infty} \frac{\sqrt{2n+1}}{n^2}$ also converges.

3. Final Answer

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

2. The series $\sum_{n=1}^{\infty} \frac{3n + 1}{n^3 - 4}$ converges.

3. The series $\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n+1}}$ converges.

4. The series $\sum_{n=1}^{\infty} \frac{\sqrt{2n+1}}{n^2}$ converges.