Let $n$ be positive integer. Define a sequence $\{a_k\}$ by 
\[a_1=\frac{1}{n(n+1)},\ a_{k+1}=-\frac{1}{k+n+1}+\frac{n}{k}\sum_{i=1}^k a_i\ \ (k=1,\ 2,\ 3,\ \cdots).\]

(1) Find $a_2$ and $a_3$.

(2) Find the general term $a_k$.

(3) Let $b_n=\sum_{k=1}^n \sqrt{a_k}$. Prove that $\lim_{n\to\infty} b_n=\ln 2$.

50 points
### Part (1): Find \( a_2 \) and \( a_3 \)

1. **Calculate \( a_2 \):**
   \[
   a_2 = -\frac{1}{n+2} + \frac{n}{1} a_1
   \]
   Given \( a_1 = \frac{1}{n(n+1)} \), substitute \( a_1 \) into the equation:
   \[
   a_2 = -\frac{1}{n+2} + n \cdot \frac{1}{n(n+1)} = -\frac{1}{n+2} + \frac{1}{n+1}
   \]
   Combine the fractions:
   \[
   a_2 = \frac{-(n+1) + (n+2)}{(n+2)(n+1)} = \frac{1}{(n+2)(n+1)}
   \]

2. **Calculate \( a_3 \):**
   \[
   a_3 = -\frac{1}{n+3} + \frac{n}{2} \sum_{i=1}^2 a_i
   \]
   Substitute \( a_1 \) and \( a_2 \):
   \[
   a_3 = -\frac{1}{n+3} + \frac{n}{2} \left( \frac{1}{n(n+1)} + \frac{1}{(n+1)(n+2)} \right)
   \]
   Combine the fractions inside the sum:
   \[
   \frac{1}{n(n+1)} + \frac{1}{(n+1)(n+2)} = \frac{(n+2) + n}{n(n+1)(n+2)} = \frac{2n+2}{n(n+1)(n+2)} = \frac{2(n+1)}{n(n+1)(n+2)} = \frac{2}{n(n+2)}
   \]
   Substitute back into the equation for \( a_3 \):
   \[
   a_3 = -\frac{1}{n+3} + \frac{n}{2} \cdot \frac{2}{n(n+2)} = -\frac{1}{n+3} + \frac{1}{n+2}
   \]
   Combine the fractions:
   \[
   a_3 = \frac{-(n+2) + (n+3)}{(n+3)(n+2)} = \frac{1}{(n+3)(n+2)}
   \]

### Part (2): Find the general term \( a_k \)

1. **Assume the general term \( a_k = \frac{1}{(n+k)(n+k-1)} \).**
2. **Prove by induction:**
   - **Base case:** For \( k = 1 \):
     \[
     a_1 = \frac{1}{n(n+1)}
     \]
     This matches the given initial condition.
   - **Inductive step:** Assume \( a_j = \frac{1}{(n+j)(n+j-1)} \) holds for some \( j \geq 1 \). Show it holds for \( j+1 \):
     \[
     a_{j+1} = -\frac{1}{n+j+1} + \frac{n}{j} \sum_{i=1}^j a_i
     \]
     Substitute the assumed form of \( a_i \):
     \[
     a_{j+1} = -\frac{1}{n+j+1} + \frac{n}{j} \sum_{i=1}^j \frac{1}{(n+i)(n+i-1)}
     \]
     The sum can be simplified using partial fractions:
     \[
     \sum_{i=1}^j \frac{1}{(n+i)(n+i-1)} = \sum_{i=1}^j \left( \frac{1}{n+i-1} - \frac{1}{n+i} \right)
     \]
     This is a telescoping series:
     \[
     \sum_{i=1}^j \left( \frac{1}{n+i-1} - \frac{1}{n+i} \right) = \frac{1}{n} - \frac{1}{n+j}
     \]
     Substitute back:
     \[
     a_{j+1} = -\frac{1}{n+j+1} + \frac{n}{j} \left( \frac{1}{n} - \frac{1}{n+j} \right) = -\frac{1}{n+j+1} + \frac{1}{j} \left( 1 - \frac{n}{n+j} \right)
     \]
     Simplify:
     \[
     a_{j+1} = -\frac{1}{n+j+1} + \frac{1}{j} \left( \frac{j}{n+j} \right) = -\frac{1}{n+j+1} + \frac{1}{n+j}
     \]
     Combine the fractions:
     \[
     a_{j+1} = \frac{-(n+j) + (n+j+1)}{(n+j+1)(n+j)} = \frac{1}{(n+j+1)(n+j)}
     \]
     Thus, the inductive step holds, and the general term is:
     \[
     a_k = \frac{1}{(n+k)(n+k-1)}
     \]

### Part (3): Prove that \( \lim_{n \to \infty} b_n = \ln 2 \)

1. **Define \( b_n \):**
   \[
   b_n = \sum_{k=1}^n \sqrt{a_k} = \sum_{k=1}^n \frac{1}{\sqrt{(n+k)(n+k-1)}}
   \]
2. **Approximate the terms:**
   \[
   \frac{1}{\sqrt{(n+k)(n+k-1)}} \approx \frac{1}{n+k}
   \]
3. **Use the integral test to approximate the sum:**
   \[
   \sum_{k=1}^n \frac{1}{n+k} \approx \int_{1}^{n} \frac{1}{n+x} \, dx
   \]
   Change of variables \( u = n + x \):
   \[
   \int_{1}^{n} \frac{1}{n+x} \, dx = \int_{n+1}^{2n} \frac{1}{u} \, du = \ln(2n) - \ln(n+1)
   \]
   As \( n \to \infty \):
   \[
   \ln(2n) - \ln(n+1) \approx \ln(2n) - \ln(n) = \ln 2
   \]
   Therefore:
   \[
   \lim_{n \to \infty} b_n = \ln 2
   \]

\(\blacksquare\)

The final answer is \( \boxed{ \ln 2 } \)