Let $a\geq 1$ be a real number. Put $x_{1}=a,x_{n+1}=1+\ln{(\frac{x_{n}^{2}}{1+\ln{x_{n}}})}(n=1,2,...)$. Prove that the sequence $\{x_{n}\}$ converges and find its limit.
1. **Define the function and sequence:**
   Let \( f(x) = 1 + \ln\left(\frac{x^2}{1 + \ln x}\right) \). The sequence is defined as \( x_1 = a \) and \( x_{n+1} = f(x_n) \) for \( n \geq 1 \).

2. **Check the derivative of \( f(x) \):**
   Compute the derivative \( f'(x) \):
   \[
   f'(x) = \frac{d}{dx} \left( 1 + \ln\left(\frac{x^2}{1 + \ln x}\right) \right)
   \]
   Using the chain rule and quotient rule:
   \[
   f'(x) = \frac{d}{dx} \left( \ln\left(\frac{x^2}{1 + \ln x}\right) \right) = \frac{1}{\frac{x^2}{1 + \ln x}} \cdot \frac{d}{dx} \left( \frac{x^2}{1 + \ln x} \right)
   \]
   Simplify the derivative inside:
   \[
   \frac{d}{dx} \left( \frac{x^2}{1 + \ln x} \right) = \frac{(1 + \ln x) \cdot 2x - x^2 \cdot \frac{1}{x}}{(1 + \ln x)^2} = \frac{2x(1 + \ln x) - x}{(1 + \ln x)^2} = \frac{x(2 + 2\ln x - 1)}{(1 + \ln x)^2} = \frac{x(1 + 2\ln x)}{(1 + \ln x)^2}
   \]
   Therefore:
   \[
   f'(x) = \frac{1 + 2\ln x}{x(1 + \ln x)}
   \]

3. **Verify the bounds of \( f'(x) \):**
   We need to show \( 0 < f'(x) \leq 1 \) for \( x \geq 1 \):
   \[
   1 + 2\ln x \leq x(1 + \ln x)
   \]
   Define \( g(x) = x(1 + \ln x) - (1 + 2\ln x) \). Compute \( g'(x) \):
   \[
   g'(x) = (1 + \ln x) + x \cdot \frac{1}{x} - \frac{2}{x} = 1 + \ln x + 1 - \frac{2}{x} = 2 + \ln x - \frac{2}{x}
   \]
   For \( x \geq 1 \), \( g'(x) \geq 0 \) because \( \ln x \geq 0 \) and \( -\frac{2}{x} \geq -2 \). Since \( g(1) = 0 \), \( g(x) \geq 0 \) for \( x \geq 1 \). Thus, \( 1 + 2\ln x \leq x(1 + \ln x) \).

4. **Show the sequence is decreasing and bounded below:**
   \[
   x_{k+1} = f(x_k) = 1 + \ln\left(\frac{x_k^2}{1 + \ln x_k}\right)
   \]
   Since \( f'(x) \leq 1 \), \( f(x) \) is non-increasing. Also, \( x_{k+1} \leq x_k \) implies the sequence is decreasing. Since \( x_k \geq 1 \), the sequence is bounded below by 1.

5. **Convergence and limit:**
   Since \( \{x_n\} \) is decreasing and bounded below, it converges to some limit \( L \). At the limit:
   \[
   L = 1 + \ln\left(\frac{L^2}{1 + \ln L}\right)
   \]
   Solve for \( L \):
   \[
   L = 1 + \ln\left(\frac{L^2}{1 + \ln L}\right)
   \]
   Assume \( L = 1 \):
   \[
   1 = 1 + \ln\left(\frac{1^2}{1 + \ln 1}\right) = 1 + \ln\left(\frac{1}{1}\right) = 1 + \ln 1 = 1
   \]
   Thus, \( L = 1 \) is a solution. Since the function \( f(x) \) is strictly decreasing for \( x \geq 1 \), the limit must be unique.

\(\blacksquare\)

The final answer is \( \boxed{ 1 } \)