Find `.^(n)C_(1)-(1)/(2).^(n)C_(2)+(1)/(3).^(n)C_(3)- . . . +(-1)^(n-1)(1)/(n).^(n)C_(n)`

Answer

To solve the problem, we need to evaluate the expression:

\[ S = \binom{n}{1} - \frac{1}{2} \binom{n}{2} + \frac{1}{3} \binom{n}{3} - \cdots + (-1)^{n-1} \frac{1}{n} \binom{n}{n} \]

### Step-by-Step Solution:

1. **Understanding the Expression**: The expression can be rewritten in summation notation as follows:
\[ S = \sum_{k=1}^{n} (-1)^{k-1} \frac{1}{k} \binom{n}{k} \]

2. **Using the Binomial Theorem**: Recall the binomial theorem states that:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
If we differentiate both sides with respect to \(x\), we get:
\[ n(1 + x)^{n-1} = \sum_{k=1}^{n} k \binom{n}{k} x^{k-1} \]

3. **Integrating**: Now, we can integrate both sides with respect to \(x\):
\[ \int n(1 + x)^{n-1} \, dx = \sum_{k=1}^{n} \binom{n}{k} \frac{x^k}{k} \]
This gives us:
\[ n \frac{(1+x)^n}{n} = \sum_{k=1}^{n} \binom{n}{k} \frac{x^k}{k} \]
Simplifying, we have:
\[ (1 + x)^n = \sum_{k=1}^{n} \binom{n}{k} \frac{x^k}{k} \]

4. **Substituting \(x = -1\)**: Now, substituting \(x = -1\) in the equation:
\[ (1 - 1)^n = \sum_{k=1}^{n} \binom{n}{k} \frac{(-1)^k}{k} \]
This simplifies to:
\[ 0 = \sum_{k=1}^{n} \binom{n}{k} \frac{(-1)^k}{k} \]

5. **Rearranging the Terms**: The sum can be rearranged to match our original expression:
\[ \sum_{k=1}^{n} (-1)^{k-1} \frac{1}{k} \binom{n}{k} = 1 \]

6. **Final Result**: Therefore, the value of \(S\) is:
\[ S = 1 \]

### Conclusion:
The final answer to the problem is:
\[ S = 1 \]