$$ \begin{array}{ll} \sum_{n=1}^{\infty} \frac{(3 n-1)}{(4 n-3)} & \sum_{n=1}^{\infty} \frac{n^{3}}{3^{n}} \ \sum_{n=1}^{\infty} \frac{1}{n!} & \sum_{n=1}^{\infty} \frac{3 n-1}{(\sqrt{2})^{n}} \ \sum_{n=1}^{\infty} \frac{1}{n} & \sum_{n=1}^{\infty} \frac{(n+2)!}{3^{n}} \ \sum_{n=1}^{\infty} \frac{(n+1)(n+2)}{n!} & \sum_{n=1}^{\infty} \frac{n\left(\frac{3}{4}\right)^{n}}{\sum_{n=1}^{\infty} \frac{n^{3}}{e^{n}}} \ \sum_{n=1}^{\infty} \frac{\sum^{n}}{n(n+2)} \frac{2^{n}}{(n-3)^{2}-2} \ \sum_{n=1}^{\infty} & \sum_{n=1}^{\infty} \frac{n^{2}+1}{n^{3}+1} \end{array} $$ D. Criterio de Cauchy o de la raiz. NOTA: Se basa en estudiar el limite de la raiz n-ésima del n-ésimo término, es decir La convergencia de la serie dependerá entonces del valor del dicho limite. - Si L 1 $$, la serie será Divergente. - Si $$ \mathrm{L}=1 $$, el criterio no da respuesta.

Question

$$ \begin{array}{ll} \sum_{n=1}^{\infty} \frac{(3 n-1)}{(4 n-3)} & \sum_{n=1}^{\infty} \frac{n^{3}}{3^{n}} \ \sum_{n=1}^{\infty} \frac{1}{n!} & \sum_{n=1}^{\infty} \frac{3 n-1}{(\sqrt{2})^{n}} \ \sum_{n=1}^{\infty} \frac{1}{n} & \sum_{n=1}^{\infty} \frac{(n+2)!}{3^{n}} \ \sum_{n=1}^{\infty} \frac{(n+1)(n+2)}{n!} & \sum_{n=1}^{\infty} \frac{n\left(\frac{3}{4}\right)^{n}}{\sum_{n=1}^{\infty} \frac{n^{3}}{e^{n}}} \ \sum_{n=1}^{\infty} \frac{\sum^{n}}{n(n+2)} \frac{2^{n}}{(n-3)^{2}-2} \ \sum_{n=1}^{\infty} & \sum_{n=1}^{\infty} \frac{n^{2}+1}{n^{3}+1} \end{array} $$ D. Criterio de Cauchy o de la raiz. NOTA: Se basa en estudiar el limite de la raiz n-ésima del n-ésimo término, es decir La convergencia de la serie dependerá entonces del valor del dicho limite. - Si L < 1, la serie será Convergente. - $$ \operatorname{Si} $$ L $$ >1 $$, la serie será Divergente. - Si $$ \mathrm{L}=1 $$, el criterio no da respuesta.

UpStudy AI · Accepted Answer

To analyze the convergence of the given series using the Cauchy root test, we will apply the following steps for each series:

1. Identify the general term of the series.
2. Calculate the $$ n $$-th root of the absolute value of the general term.
3. Take the limit as $$ n $$ approaches infinity.
4. Determine convergence or divergence based on the limit $$ L $$.

Let's analyze each series one by one.

### 1. $$ \sum_{n=1}^{\infty} \frac{(3 n-1)}{(4 n-3)} $$

**General term:** $$ a_n = \frac{3n-1}{4n-3} $$

**Limit calculation:**
$$
L = \lim_{n \to \infty} \sqrt[n]{\left| a_n \right|} = \lim_{n \to \infty} \sqrt[n]{\frac{3n-1}{4n-3}} = \lim_{n \to \infty} \frac{3n-1}{4n-3} = \frac{3}{4}
$$

Since $$ L < 1 $$, the series converges.

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

**General term:** $$ a_n = \frac{n^3}{3^n} $$

**Limit calculation:**
$$
L = \lim_{n \to \infty} \sqrt[n]{\left| a_n \right|} = \lim_{n \to \infty} \sqrt[n]{\frac{n^3}{3^n}} = \lim_{n \to \infty} \frac{n^{3/n}}{3} = \frac{1}{3}
$$

Since $$ L < 1 $$, the series converges.

### 3. $$ \sum_{n=1}^{\infty} \frac{1}{n!} $$

**General term:** $$ a_n = \frac{1}{n!} $$

**Limit calculation:**
$$
L = \lim_{n \to \infty} \sqrt[n]{\left| a_n \right|} = \lim_{n \to \infty} \sqrt[n]{\frac{1}{n!}} = 0
$$

Since $$ L < 1 $$, the series converges.

### 4. $$ \sum_{n=1}^{\infty} \frac{3 n-1}{(\sqrt{2})^{n}} $$

**General term:** $$ a_n = \frac{3n-1}{(\sqrt{2})^n} $$

**Limit calculation:**
$$
L = \lim_{n \to \infty} \sqrt[n]{\left| a_n \right|} = \lim_{n \to \infty} \frac{3n-1}{(\sqrt{2})^n} = \frac{1}{\sqrt{2}} < 1
$$

Since $$ L < 1 $$, the series converges.

### 5. $$ \sum_{n=1}^{\infty} \frac{1}{n} $$

**General term:** $$ a_n = \frac{1}{n} $$

**Limit calculation:**
$$
L = \lim_{n \to \infty} \sqrt[n]{\left| a_n \right|} = \lim_{n \to \infty} \sqrt[n]{\frac{1}{n}} = 1
$$

Since $$ L = 1 $$, the test is inconclusive.

### 6. $$ \sum_{n=1}^{\infty} \frac{(n+2)!}{3^{n}} $$

**General term:** $$ a_n = \frac{(n+2)!}{3^n} $$

**Limit calculation:**
$$
L = \lim_{n \to \infty} \sqrt[n]{\left| a_n \right|} = \lim_{n \to \infty} \frac{(n+2)!^{1/n}}{3} \approx \frac{n}{3} \to \infty
$$

Since $$ L > 1 $$, the series diverges.

### 7. $$ \sum_{n=1}^{\infty} \frac{(n+1)(n+2)}{n!} $$

**General term:** $$ a_n = \frac{(n+1)(n+2)}{n!} $$

**Limit calculation:**
$$
L = \lim_{n \to \infty} \sqrt[n]{\left| a_n \right|} = \lim_{n \to \infty} \frac{(n+1)(n+2)^{1/n}}{n!^{1/n}} = 0
$$

Since $$ L < 1 $$, the series converges.

### 8. $$ \sum_{n=1}^{\infty} \frac{n\left(\frac{3}{4}\right)^{n}}{\sum_{n=1}^{\infty} \frac{n^{3}}{e^{n}}} $$

**General term:** $$ a_n = \frac{n\left(\frac{3}{4}\right)^{n}}{C} $$ where $$ C = \sum_{n=1}^{\infty} \frac{n^{3}}{e^{n}} $$ (which converges).

**Limit calculation:**
$$
L = \lim_{n \to \infty} \sqrt[n]{\left| a_n \right|} = \lim_{n \to \infty} \frac{n^{1/n} \cdot \left(\frac{3}{4}\right)}{C^{1/n}} = \frac{3}{4}
$$

Since $$ L < 1 $$, the series converges.

### 9. $$ \sum_{n=1}^{\infty} \frac{\sum^{n}}{n(n+2)} $$

This series needs clarification on the inner summation. Assuming it converges, we can analyze it similarly.

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

**General term:** $$ a_n = \frac{n^2 + 1}{n^3 + 1} $$

**Limit calculation:**
$$
L = \lim_{n \to \infty} \sqrt[n]{\left| a_n \right|} = \lim_{n \to \infty} \frac{n^2 + 1}{n^3 + 1} = 0
$$

Since $$ L < 1 $$, the series converges.

### Summary of Results:
- Convergent: $$
- Convergent: Series 1, 2, 3, 4, 7, 8, 10 - Divergent: Series 6 - Inconclusive: Series 5

Upstudy AI Solution

Answer

Solution

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Summary of Results:

The Deep Dive

Related Questions

Latest Calculus Questions