Let `s_(n)=1+(1)/(3)+(1)/(3^(2))+…+(1)/(3^(n-1))`. The least value of `n in N` such that `(3)/(2) -S_(n) lt (1)/(400)` is

Answer

To solve the problem, we need to find the least value of \( n \in \mathbb{N} \) such that

\[ \frac{3}{2} - S_n < \frac{1}{400} \]

where

\[ S_n = 1 + \frac{1}{3} + \frac{1}{3^2} + \ldots + \frac{1}{3^{n-1}}. \]

### Step 1: Identify the series as a geometric progression (GP)

The series \( S_n \) is a geometric progression where:
- The first term \( a = 1 \)
- The common ratio \( r = \frac{1}{3} \)

### Step 2: Use the formula for the sum of the first \( n \) terms of a GP

The sum of the first \( n \) terms of a GP is given by:

\[ S_n = \frac{a(1 - r^n)}{1 - r} \]

Substituting the values of \( a \) and \( r \):

\[ S_n = \frac{1(1 - (\frac{1}{3})^n)}{1 - \frac{1}{3}} = \frac{1 - \frac{1}{3^n}}{\frac{2}{3}} = \frac{3}{2}(1 - \frac{1}{3^n}) = \frac{3}{2} - \frac{3}{2 \cdot 3^n} \]

### Step 3: Substitute \( S_n \) into the inequality

Now, substituting \( S_n \) back into the inequality:

\[ \frac{3}{2} - \left( \frac{3}{2} - \frac{3}{2 \cdot 3^n} \right) < \frac{1}{400} \]

This simplifies to:

\[ \frac{3}{2 \cdot 3^n} < \frac{1}{400} \]

### Step 4: Rearranging the inequality

Multiplying both sides by \( 400 \cdot 2 \cdot 3^n \) (which is positive for \( n \in \mathbb{N} \)) gives:

\[ 3 \cdot 400 < 2 \cdot 3^n \]

This simplifies to:

\[ 1200 < 2 \cdot 3^n \]

Dividing both sides by 2:

\[ 600 < 3^n \]

### Step 5: Finding the least value of \( n \)

Now we need to find the smallest integer \( n \) such that \( 3^n > 600 \).

Calculating powers of 3:

- \( 3^1 = 3 \)
- \( 3^2 = 9 \)
- \( 3^3 = 27 \)
- \( 3^4 = 81 \)
- \( 3^5 = 243 \)
- \( 3^6 = 729 \)

We see that \( 3^5 = 243 \) is less than 600, and \( 3^6 = 729 \) is greater than 600. Therefore, the least value of \( n \) is:

\[ n = 6 \]

### Final Answer

The least value of \( n \in \mathbb{N} \) such that \( \frac{3}{2} - S_n < \frac{1}{400} \) is

\[ \boxed{6} \]