Answer the following questions.

(1) $ 0 < x\leq 2\pi$, prove that $ |\sin x| < x$.

(2) Let $ f_1(x) \equal{} \sin x\ , a$ be the constant such that $ 0 < a\leq 2\pi$.

Define $ f_{n \plus{} 1}(x) \equal{} \frac {1}{2a}\int_{x \minus{} a}^{x \plus{} a} f_n(t)\ dt\ (n \equal{} 1,\ 2,\ 3,\ \cdots)$. Find $ f_2(x)$.

(3) Find $ f_n(x)$ for all $ n$.

(4) For a given $ x$, find $ \sum_{n \equal{} 1}^{\infty} f_n(x)$.
1. **Prove that \( |\sin x| < x \) for \( 0 < x \leq 2\pi \):**

   To prove \( |\sin x| < x \) for \( 0 < x \leq 2\pi \), we can consider the behavior of the sine function and the line \( y = x \).

   - For \( 0 < x \leq \frac{\pi}{2} \):
     - The function \( \sin x \) is increasing and \( \sin x \leq 1 \).
     - Since \( x \leq \frac{\pi}{2} \approx 1.57 \), it follows that \( \sin x < x \).

   - For \( \frac{\pi}{2} < x \leq \pi \):
     - The function \( \sin x \) is decreasing and \( \sin x \leq 1 \).
     - Since \( x > \frac{\pi}{2} \), it follows that \( \sin x < x \).

   - For \( \pi < x \leq 2\pi \):
     - The function \( \sin x \) is negative, and \( |\sin x| \leq 1 \).
     - Since \( x > \pi \), it follows that \( |\sin x| < x \).

   Therefore, \( |\sin x| < x \) for \( 0 < x \leq 2\pi \).

   \(\blacksquare\)

2. **Find \( f_2(x) \):**

   Given:
   \[
   f_{n+1}(x) = \frac{1}{2a} \int_{x-a}^{x+a} f_n(t) \, dt
   \]
   and \( f_1(x) = \sin x \).

   To find \( f_2(x) \):
   \[
   f_2(x) = \frac{1}{2a} \int_{x-a}^{x+a} \sin t \, dt
   \]

   Using the integral of sine:
   \[
   \int \sin t \, dt = -\cos t
   \]

   Therefore:
   \[
   f_2(x) = \frac{1}{2a} \left[ -\cos t \right]_{x-a}^{x+a} = \frac{1}{2a} \left[ -\cos(x+a) + \cos(x-a) \right]
   \]

   Using the trigonometric identity for the difference of cosines:
   \[
   \cos(x-a) - \cos(x+a) = -2 \sin x \sin a
   \]

   Thus:
   \[
   f_2(x) = \frac{1}{2a} \left[ -(-2 \sin x \sin a) \right] = \frac{1}{2a} \cdot 2 \sin x \sin a = \frac{\sin a}{a} \sin x
   \]

3. **Find \( f_n(x) \) for all \( n \):**

   By induction, assume \( f_n(x) = \left( \frac{\sin a}{a} \right)^{n-1} \sin x \).

   For \( n = 1 \):
   \[
   f_1(x) = \sin x
   \]

   Assume it holds for \( n \):
   \[
   f_n(x) = \left( \frac{\sin a}{a} \right)^{n-1} \sin x
   \]

   Then for \( n+1 \):
   \[
   f_{n+1}(x) = \frac{1}{2a} \int_{x-a}^{x+a} f_n(t) \, dt = \frac{1}{2a} \int_{x-a}^{x+a} \left( \frac{\sin a}{a} \right)^{n-1} \sin t \, dt
   \]

   Using the result from \( f_2(x) \):
   \[
   f_{n+1}(x) = \left( \frac{\sin a}{a} \right)^{n-1} \frac{\sin a}{a} \sin x = \left( \frac{\sin a}{a} \right)^n \sin x
   \]

   Therefore:
   \[
   f_n(x) = \left( \frac{\sin a}{a} \right)^{n-1} \sin x
   \]

4. **Find \( \sum_{n=1}^{\infty} f_n(x) \):**

   Using the result from (3):
   \[
   \sum_{n=1}^{\infty} f_n(x) = \sum_{n=1}^{\infty} \left( \frac{\sin a}{a} \right)^{n-1} \sin x
   \]

   This is a geometric series with the first term \( \sin x \) and common ratio \( \frac{\sin a}{a} \):
   \[
   \sum_{n=1}^{\infty} \left( \frac{\sin a}{a} \right)^{n-1} = \frac{1}{1 - \frac{\sin a}{a}} = \frac{a}{a - \sin a}
   \]

   Therefore:
   \[
   \sum_{n=1}^{\infty} f_n(x) = \sin x \cdot \frac{a}{a - \sin a} = \frac{a \sin x}{a - \sin a}
   \]

The final answer is \( \boxed{ \frac{a \sin x}{a - \sin a} } \)