Number of positive terms in the sequence `x_n=195/(4P_n)-(n+3p_3)/(P_(n+1)), n in N` (here `p_n=|anglen`)

Explanation

To solve the problem, we need to analyze the sequence given by

\[ x_n = \frac{195}{4P_n} - \frac{(n + 3)P_3}{P_{n+1}} \]

where \( P_n = n! \) (the factorial of \( n \)).

### Step-by-Step Solution:

1. **Substituting \( P_n \) and \( P_{n+1} \)**:
We know that \( P_n = n! \) and \( P_{n+1} = (n+1)! = (n+1) \cdot n! \). Thus, we can rewrite the sequence:

\[ x_n = \frac{195}{4n!} - \frac{(n + 3) \cdot 6}{(n + 1) \cdot n!} \]

Here, \( P_3 = 3! = 6 \).

2. **Combining the terms**:
We can combine the two fractions by finding a common denominator, which is \( 4n!(n + 1) \):

\[ x_n = \frac{195(n + 1) - 4(n + 3) \cdot 6}{4n!(n + 1)} \]

Simplifying the numerator:

\[ = \frac{195(n + 1) - 24(n + 3)}{4n!(n + 1)} \]

\[ = \frac{195n + 195 - 24n - 72}{4n!(n + 1)} \]

\[ = \frac{171n + 123}{4n!(n + 1)} \]

3. **Finding when \( x_n > 0 \)**:
For \( x_n \) to be positive, the numerator must be greater than zero:

\[ 171n + 123 > 0 \]

Solving for \( n \):

\[ 171n > -123 \]

\[ n > -\frac{123}{171} \approx -0.72 \]

Since \( n \) is a natural number, this condition is satisfied for all positive integers.

4. **Checking the denominator**:
The denominator \( 4n!(n + 1) \) is always positive for \( n \in \mathbb{N} \).

5. **Finding the upper limit**:
We need to determine if there is an upper limit for \( n \). We can analyze the quadratic formed from the numerator:

\[ 171n + 123 = 0 \]

This does not provide an upper limit since it is always positive for \( n \geq 1 \).

6. **Conclusion**:
The sequence \( x_n \) is positive for all \( n \) in \( \mathbb{N} \). Therefore, we need to find the number of positive integers \( n \) such that \( x_n \) remains positive.

Since there is no upper limit derived from the conditions, we conclude that there are infinitely many positive terms in the sequence.

### Final Answer:
The number of positive terms in the sequence is infinite.