When `0 lt x lt 1 , y=1/2x^2+2/3x^3+3/4x^4+ . . . ` Then the value of `e^(1-y)` at `x=1/2`

Answer

To solve the problem, we need to find the value of \( e^{1-y} \) at \( x = \frac{1}{2} \), where

\[ y = \frac{1}{2} x^2 + \frac{2}{3} x^3 + \frac{3}{4} x^4 + \ldots \]

### Step 1: Express \( y \) in a more manageable form

We can observe that the series for \( y \) can be rewritten as:

\[ y = \sum_{n=2}^{\infty} \frac{n-1}{n} x^n \]

This can be separated into two parts:

\[ y = \sum_{n=2}^{\infty} x^n - \sum_{n=2}^{\infty} \frac{x^n}{n} \]

### Step 2: Calculate the first sum

The first sum, \( \sum_{n=2}^{\infty} x^n \), is a geometric series:

\[ \sum_{n=2}^{\infty} x^n = x^2 + x^3 + x^4 + \ldots = \frac{x^2}{1-x} \quad \text{for } 0 < x < 1 \]

### Step 3: Calculate the second sum

The second sum, \( \sum_{n=2}^{\infty} \frac{x^n}{n} \), can be recognized as part of the Taylor series expansion for \( -\ln(1-x) \):

\[ \sum_{n=1}^{\infty} \frac{x^n}{n} = -\ln(1-x) \quad \Rightarrow \quad \sum_{n=2}^{\infty} \frac{x^n}{n} = -\ln(1-x) - x \]

### Step 4: Combine the results

Now we can combine the two parts:

\[ y = \frac{x^2}{1-x} + \ln(1-x) + x \]

### Step 5: Substitute \( x = \frac{1}{2} \)

Now, we substitute \( x = \frac{1}{2} \):

\[ y = \frac{\left(\frac{1}{2}\right)^2}{1 - \frac{1}{2}} + \ln\left(1 - \frac{1}{2}\right) + \frac{1}{2} \]

Calculating each term:

1. \( \frac{\left(\frac{1}{2}\right)^2}{1 - \frac{1}{2}} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} \)
2. \( \ln\left(1 - \frac{1}{2}\right) = \ln\left(\frac{1}{2}\right) = -\ln(2) \)
3. \( \frac{1}{2} \)

So,

\[ y = \frac{1}{2} - \ln(2) + \frac{1}{2} = 1 - \ln(2) \]

### Step 6: Calculate \( e^{1-y} \)

Now, we need to find \( e^{1-y} \):

\[ 1 - y = 1 - (1 - \ln(2)) = \ln(2) \]

Thus,

\[ e^{1-y} = e^{\ln(2)} = 2 \]

### Final Answer

The value of \( e^{1-y} \) at \( x = \frac{1}{2} \) is \( \boxed{2} \).