If `x gt 0`, `(x^(n))/(1+x+x^(2)+...+x^(2n))` is

Options

A. ` le (1)/(2n+1)`
B. ` lt (2)/(2n+1)`
C. ` ge (1)/(2n+1)`
D. ` gt (2)/(2n+1)`

Explanation

To solve the problem, we need to analyze the expression given:

\[ \frac{x^n}{1 + x + x^2 + \ldots + x^{2n}} \]

### Step 1: Simplifying the Denominator

The denominator is a geometric series. We can use the formula for the sum of a geometric series:

\[ S = a \frac{1 - r^{n}}{1 - r} \]

where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. In our case:

- The first term \( a = 1 \)
- The common ratio \( r = x \)
- The number of terms is \( 2n + 1 \) (from \( x^0 \) to \( x^{2n} \))

Thus, the sum of the series is:

\[ 1 + x + x^2 + \ldots + x^{2n} = \frac{1 - x^{2n+1}}{1 - x} \]

### Step 2: Rewrite the Expression

Now we can rewrite our expression:

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

### Step 3: Analyzing the Expression

We need to analyze the behavior of this expression as \( x > 0 \).

1. **As \( x \to 0 \)**:
\[ \frac{x^n (1 - x)}{1 - x^{2n+1}} \to 0 \]

2. **As \( x \to 1 \)**:
\[ \frac{x^n (1 - x)}{1 - x^{2n+1}} \to \frac{1 \cdot 0}{0} \text{ (indeterminate form)} \]

3. **As \( x \to \infty \)**:
\[ \frac{x^n (1 - x)}{1 - x^{2n+1}} \to \text{ (dominant terms lead to } \frac{x^n (-x)}{-x^{2n+1}} = \frac{x^{n+1}}{x^{2n+1}} = \frac{1}{x^{n}} \to 0) \]

### Step 4: Finding the Maximum Value

To find the maximum value of the expression, we can apply the AM-GM inequality.

We know that:

\[ \frac{x^k + \frac{1}{x^k}}{2} \geq 1 \quad \text{for } k = 1, 2, \ldots, n \]

This implies:

\[ x + \frac{1}{x} \geq 2, \quad x^2 + \frac{1}{x^2} \geq 2, \ldots, \quad x^n + \frac{1}{x^n} \geq 2 \]

Adding these inequalities gives:

\[ (x + \frac{1}{x}) + (x^2 + \frac{1}{x^2}) + \ldots + (x^n + \frac{1}{x^n}) \geq 2n \]

Thus:

\[ 1 + x + x^2 + \ldots + x^{2n} \geq 2n + 1 \]

Taking the reciprocal gives:

\[ \frac{1}{1 + x + x^2 + \ldots + x^{2n}} \leq \frac{1}{2n + 1} \]

### Conclusion

Thus, we conclude that:

\[ \frac{x^n}{1 + x + x^2 + \ldots + x^{2n}} \leq \frac{1}{2n + 1} \]

The maximum value of the expression is:

\[ \frac{1}{2n + 1} \]

### Final Answer

The answer is:

\[ \frac{x^n}{1 + x + x^2 + \ldots + x^{2n}} \leq \frac{1}{2n + 1} \]