The task requires determining the convergence of the series using Cauchy's root test (радикальный признак Коши). The series provided is:

$\sum_{n=1}^\infty (-1)^{n-1} \left( \frac{3n + 1}{4n + 7} \right)^{2n}$

Step-by-Step Solution

The Cauchy root test examines the limit:

$\lim_{n \to \infty} \sqrt[n]{|a_n|},$ where $a_n$ is the $n$-th term of the series.

• If $L < 1$, the series converges absolutely.
• If $L > 1$, the series diverges.
• If $L = 1$, the test is inconclusive.

1. Extract the $n$-th term $a_n$:

The term of the series is:

$a_n = \left( \frac{3n + 1}{4n + 7} \right)^{2n}.$

2. Apply the root test:

We calculate the $n$-th root of $|a_n|$:

[ \sqrt[n]{|a_n|} = \sqrt[n]{\left( \frac{3n + 1}{4n + 7} \right)^{2n}}. ]

Simplifying the expression:

$\sqrt[n]{\left( \frac{3n + 1}{4n + 7} \right)^{2n}} = \left( \frac{3n + 1}{4n + 7} \right)^2.$

3. Find the limit as $n \to \infty$:

To compute the limit:

$\lim_{n \to \infty} \left( \frac{3n + 1}{4n + 7} \right)^2.$

• For large $n$, the dominant terms in the numerator and denominator are $3n$ and $4n$, respectively.
• Therefore:

$\frac{3n + 1}{4n + 7} \sim \frac{3n}{4n} = \frac{3}{4}.$

Taking the square:

$\left( \frac{3}{4} \right)^2 = \frac{9}{16}.$

Thus:

$\lim_{n \to \infty} \sqrt[n]{|a_n|} = \frac{9}{16}.$

4. Interpret the result:

Since $\frac{9}{16} < 1$, the series converges absolutely.

• Result to input: 1.52
  (1 for absolute convergence and 52 for the result $0.52$ relative to the example format).

Final Answer:

The series converges absolutely, so the result is 1.52.

Do you need further clarification or additional details?

Here are 5 related questions for practice:

1. What is the general form of Cauchy's root test?
2. How can you determine whether a series converges conditionally instead of absolutely?
3. How does the comparison test help when $L = 1$ in the root test?
4. What is the difference between Cauchy's root test and the ratio test?
5. What happens if the root test gives a limit exactly equal to 1?

Tip: Always simplify the dominant terms in limits for large $n$ to find approximations efficiently.