Let `a lt b lt 0` and `I(n) =a^(1//n)-b^(1//n),J(n)=(a-b)^(1//n)` for all `n ge 2` then

Options

A. `I(n) lt j(n)`
B. `I(n) gt j(n)`
C. `I(n)=j(n)`
D. `I(n)+j(n)=0`

Answer

To solve the problem, we need to analyze the expressions \( I(n) \) and \( J(n) \) given the conditions \( a < b < 0 \).

### Step-by-Step Solution:

1. **Understanding the Expressions**:
- We have \( I(n) = a^{1/n} - b^{1/n} \)
- We also have \( J(n) = (a - b)^{1/n} \)

2. **Analyzing the Values of \( a \) and \( b \)**:
- Since \( a < b < 0 \), both \( a \) and \( b \) are negative numbers.
- This means \( a - b < 0 \) because \( a < b \).

3. **Rewriting the Expressions**:
- Since \( a \) and \( b \) are negative, we can express \( a \) and \( b \) as \( a = -m \) and \( b = -n \) where \( m, n > 0 \) and \( m > n \).
- Thus, \( I(n) = (-m)^{1/n} - (-n)^{1/n} \) and \( J(n) = (-m + n)^{1/n} \).

4. **Finding the Ratio \( \frac{I(n)}{J(n)} \)**:
- We will compute \( \frac{I(n)}{J(n)} \):
\[ \frac{I(n)}{J(n)} = \frac{(-m)^{1/n} - (-n)^{1/n}}{(-m + n)^{1/n}} \]

5. **Simplifying the Ratio**:
- We can factor out \( (-n)^{1/n} \) from the numerator:
\[ = \frac{(-n)^{1/n} \left( \left( \frac{-m}{-n} \right)^{1/n} - 1 \right)}{(-m + n)^{1/n}} \]
- Let \( \frac{-m}{-n} = \frac{m}{n} \), then:
\[ = \frac{(-n)^{1/n} \left( \left( \frac{m}{n} \right)^{1/n} - 1 \right)}{(-m + n)^{1/n}} \]

6. **Behavior as \( n \to \infty \)**:
- As \( n \) increases, both \( (-m)^{1/n} \) and \( (-n)^{1/n} \) approach 0.
- The term \( \left( \frac{m}{n} \right)^{1/n} \) approaches 1 as \( n \to \infty \).

7. **Conclusion**:
- Therefore, \( I(n) \) becomes less than \( J(n) \) as \( n \) increases, leading to:
\[ \frac{I(n)}{J(n)} < 1 \]

### Final Result:
Thus, we conclude that \( I(n) < J(n) \) for all \( n \geq 2 \).

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