Consider a right-angled triangle with $AB=1,\ AC=\sqrt{3},\ \angle{BAC}=\frac{\pi}{2}.$ Let $P_1,\ P_2,\ \cdots\cdots,\ P_{n-1}\ (n\geq 2)$ be the points  which are closest from $A$, in this order and obtained by dividing $n$ equally parts of the line segment $AB$. Denote by $A=P_0,\ B=P_n$, answer the questions as below.

(1) Find the inradius of $\triangle{P_kCP_{k+1}}\ (0\leq k\leq n-1)$.

(2) Denote by $S_n$ the total sum of the area of the incircle for $\triangle{P_kCP_{k+1}}\ (0\leq k\leq n-1)$.

Let $I_n=\frac{1}{n}\sum_{k=0}^{n-1} \frac{1}{3+\left(\frac{k}{n}\right)^2}$, show that $nS_n\leq \frac {3\pi}4I_n$, then find the limit $\lim_{n\to\infty} I_n$.

(3) Find the limit $\lim_{n\to\infty} nS_n$.
Given a right-angled triangle \( \triangle ABC \) with \( AB = 1 \), \( AC = \sqrt{3} \), and \( \angle BAC = \frac{\pi}{2} \). Points \( P_1, P_2, \ldots, P_{n-1} \) divide \( AB \) into \( n \) equal parts. We need to find the inradius of \( \triangle P_kCP_{k+1} \) and the total sum of the areas of the incircles for these triangles.

1. **Finding the inradius of \( \triangle P_kCP_{k+1} \):**

   - The length of each segment \( P_kP_{k+1} \) is \( \frac{1}{n} \).
   - The coordinates of \( P_k \) are \( \left(\frac{k}{n}, 0\right) \) and \( P_{k+1} \) are \( \left(\frac{k+1}{n}, 0\right) \).
   - The triangle \( \triangle P_kCP_{k+1} \) is a right-angled triangle with \( P_kP_{k+1} \) as the base and \( C \) as the vertex opposite the right angle.

   Using the formula for the inradius \( r \) of a right-angled triangle with legs \( a \) and \( b \) and hypotenuse \( c \):
   \[
   r = \frac{a + b - c}{2}
   \]
   Here, \( a = P_kP_{k+1} = \frac{1}{n} \), \( b = CP_k \), and \( c = CP_{k+1} \).

   - The length \( CP_k \) can be found using the distance formula:
     \[
     CP_k = \sqrt{\left(\frac{k}{n}\right)^2 + (\sqrt{3})^2} = \sqrt{\left(\frac{k}{n}\right)^2 + 3}
     \]
   - Similarly,
     \[
     CP_{k+1} = \sqrt{\left(\frac{k+1}{n}\right)^2 + 3}
     \]

   Since \( n \) is large, the difference between \( CP_k \) and \( CP_{k+1} \) is small, and we can approximate:
   \[
   CP_k \approx CP_{k+1} \approx \sqrt{\left(\frac{k}{n}\right)^2 + 3}
   \]

   Therefore, the inradius \( r \) is approximately:
   \[
   r \approx \frac{\frac{1}{n} + \sqrt{\left(\frac{k}{n}\right)^2 + 3} - \sqrt{\left(\frac{k}{n}\right)^2 + 3}}{2} = \frac{1}{2n}
   \]

2. **Finding \( S_n \), the total sum of the areas of the incircles:**

   The area of the incircle of \( \triangle P_kCP_{k+1} \) is:
   \[
   \text{Area} = \pi r^2 = \pi \left(\frac{1}{2n}\right)^2 = \frac{\pi}{4n^2}
   \]

   Summing this for all \( k \) from 0 to \( n-1 \):
   \[
   S_n = \sum_{k=0}^{n-1} \frac{\pi}{4n^2} = \frac{n \cdot \frac{\pi}{4n^2}}{n} = \frac{\pi}{4n}
   \]

3. **Finding the limit \( \lim_{n \to \infty} I_n \):**

   Given:
   \[
   I_n = \frac{1}{n} \sum_{k=0}^{n-1} \frac{1}{3 + \left(\frac{k}{n}\right)^2}
   \]

   As \( n \to \infty \), the sum can be approximated by an integral:
   \[
   I_n \approx \int_0^1 \frac{1}{3 + x^2} \, dx
   \]

   Evaluating the integral:
   \[
   \int_0^1 \frac{1}{3 + x^2} \, dx = \left. \frac{1}{\sqrt{3}} \arctan\left(\frac{x}{\sqrt{3}}\right) \right|_0^1 = \frac{1}{\sqrt{3}} \left( \arctan\left(\frac{1}{\sqrt{3}}\right) - \arctan(0) \right) = \frac{1}{\sqrt{3}} \cdot \frac{\pi}{6} = \frac{\pi}{6\sqrt{3}}
   \]

   Therefore,
   \[
   \lim_{n \to \infty} I_n = \frac{\pi}{6\sqrt{3}}
   \]

4. **Finding the limit \( \lim_{n \to \infty} nS_n \):**

   From the previous calculations:
   \[
   nS_n = n \cdot \frac{\pi}{4n} = \frac{\pi}{4}
   \]

   Therefore,
   \[
   \lim_{n \to \infty} nS_n = \frac{\pi}{4}
   \]

The final answer is \(\boxed{\frac{\pi}{4}}\)