Which of the following rationals is greater ?

$\frac{1995^{1994} + 1}{1995^{1995} + 1}$ or $\frac{1995^{1995} + 1}{ 1995^{1996} +1}$
1. Let's denote \( x = 1995 \). We need to compare the two fractions:
   \[
   \frac{x^{1994} + 1}{x^{1995} + 1} \quad \text{and} \quad \frac{x^{1995} + 1}{x^{1996} + 1}
   \]

2. Consider the function \( f(x) = \frac{x^n + 1}{x^{n+1} + 1} \). We need to show that this function is decreasing for \( x > 0 \).

3. To show that \( f(x) \) is decreasing, we can compute its derivative and show that it is negative for \( x > 0 \).

4. The derivative of \( f(x) \) with respect to \( x \) is:
   \[
   f'(x) = \frac{d}{dx} \left( \frac{x^n + 1}{x^{n+1} + 1} \right)
   \]

5. Using the quotient rule, we have:
   \[
   f'(x) = \frac{(x^{n+1} + 1) \cdot \frac{d}{dx}(x^n + 1) - (x^n + 1) \cdot \frac{d}{dx}(x^{n+1} + 1)}{(x^{n+1} + 1)^2}
   \]

6. Simplifying the derivatives:
   \[
   \frac{d}{dx}(x^n + 1) = n x^{n-1}
   \]
   \[
   \frac{d}{dx}(x^{n+1} + 1) = (n+1) x^n
   \]

7. Substituting these into the derivative expression:
   \[
   f'(x) = \frac{(x^{n+1} + 1) \cdot n x^{n-1} - (x^n + 1) \cdot (n+1) x^n}{(x^{n+1} + 1)^2}
   \]

8. Simplifying the numerator:
   \[
   f'(x) = \frac{n x^{n-1} x^{n+1} + n x^{n-1} - (n+1) x^n x^n - (n+1) x^n}{(x^{n+1} + 1)^2}
   \]
   \[
   f'(x) = \frac{n x^{2n} + n x^{n-1} - (n+1) x^{2n} - (n+1) x^n}{(x^{n+1} + 1)^2}
   \]
   \[
   f'(x) = \frac{n x^{2n} + n x^{n-1} - n x^{2n} - x^{2n} - (n+1) x^n}{(x^{n+1} + 1)^2}
   \]
   \[
   f'(x) = \frac{- x^{2n} + n x^{n-1} - (n+1) x^n}{(x^{n+1} + 1)^2}
   \]

9. Since \( x > 0 \), the term \( - x^{2n} \) dominates, making \( f'(x) < 0 \). Therefore, \( f(x) \) is a decreasing function for \( x > 0 \).

10. Applying this result to our specific case with \( x = 1995 \), \( n = 1994 \), and \( n+1 = 1995 \):
    \[
    \frac{1995^{1994} + 1}{1995^{1995} + 1} > \frac{1995^{1995} + 1}{1995^{1996} + 1}
    \]

Conclusion:
\[
\boxed{\frac{1995^{1994} + 1}{1995^{1995} + 1} > \frac{1995^{1995} + 1}{1995^{1996} + 1}}
\]