Let $$a_n = \frac{1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}.$$
(a) Prove that $\lim_{n\to \infty}a_n$ exists.
(b) Show that $$a_n = \frac{\left(1-\frac1{2^2}\right)\left(1-\frac1{4^2}\right)\left(1-\frac1{6^2}\right)\cdots\left(1-\frac{1}{(2n)^2}\right)}{(2n+1)a_n}.$$
(c) Find $\lim_{n\to\infty}a_n$ and justify your answer
### Part (a): Prove that $\lim_{n\to \infty}a_n$ exists.

1. **Expression for \(a_n\):**
   \[
   a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n}
   \]

2. **Simplify the product:**
   \[
   a_n = \prod_{k=1}^n \frac{2k-1}{2k}
   \]

3. **Rewrite each term:**
   \[
   \frac{2k-1}{2k} = 1 - \frac{1}{2k}
   \]

4. **Product form:**
   \[
   a_n = \prod_{k=1}^n \left(1 - \frac{1}{2k}\right)
   \]

5. **Logarithm of the product:**
   \[
   \ln(a_n) = \sum_{k=1}^n \ln\left(1 - \frac{1}{2k}\right)
   \]

6. **Approximate the logarithm for small \(x\):**
   \[
   \ln(1 - x) \approx -x \quad \text{for small } x
   \]

7. **Apply the approximation:**
   \[
   \ln(a_n) \approx \sum_{k=1}^n -\frac{1}{2k} = -\frac{1}{2} \sum_{k=1}^n \frac{1}{k}
   \]

8. **Harmonic series approximation:**
   \[
   \sum_{k=1}^n \frac{1}{k} \approx \ln(n) + \gamma \quad \text{(where \(\gamma\) is the Euler-Mascheroni constant)}
   \]

9. **Substitute the harmonic series approximation:**
   \[
   \ln(a_n) \approx -\frac{1}{2} (\ln(n) + \gamma)
   \]

10. **Exponentiate to find \(a_n\):**
    \[
    a_n \approx e^{-\frac{1}{2} (\ln(n) + \gamma)} = \frac{e^{-\frac{\gamma}{2}}}{\sqrt{n}}
    \]

11. **Limit of \(a_n\):**
    \[
    \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{e^{-\frac{\gamma}{2}}}{\sqrt{n}} = 0
    \]

Thus, the limit exists and is 0.

### Part (b): Show that 
\[ 
a_n = \frac{\left(1-\frac1{2^2}\right)\left(1-\frac1{4^2}\right)\left(1-\frac1{6^2}\right)\cdots\left(1-\frac{1}{(2n)^2}\right)}{(2n+1)a_n}.
\]

1. **Rewrite \(a_n\):**
   \[
   a_n = \prod_{k=1}^n \left(1 - \frac{1}{2k}\right)
   \]

2. **Consider the product in the numerator:**
   \[
   \prod_{k=1}^n \left(1 - \frac{1}{(2k)^2}\right)
   \]

3. **Simplify the product:**
   \[
   \left(1 - \frac{1}{2^2}\right)\left(1 - \frac{1}{4^2}\right)\left(1 - \frac{1}{6^2}\right) \cdots \left(1 - \frac{1}{(2n)^2}\right)
   \]

4. **Combine the terms:**
   \[
   \prod_{k=1}^n \left(1 - \frac{1}{(2k)^2}\right) = \prod_{k=1}^n \left(\frac{(2k)^2 - 1}{(2k)^2}\right) = \prod_{k=1}^n \left(\frac{4k^2 - 1}{4k^2}\right)
   \]

5. **Factorize the numerator:**
   \[
   \prod_{k=1}^n \left(\frac{(2k-1)(2k+1)}{4k^2}\right) = \prod_{k=1}^n \left(\frac{2k-1}{2k}\right) \cdot \prod_{k=1}^n \left(\frac{2k+1}{2k}\right)
   \]

6. **Separate the products:**
   \[
   \prod_{k=1}^n \left(\frac{2k-1}{2k}\right) \cdot \prod_{k=1}^n \left(\frac{2k+1}{2k}\right) = a_n \cdot \prod_{k=1}^n \left(\frac{2k+1}{2k}\right)
   \]

7. **Simplify the second product:**
   \[
   \prod_{k=1}^n \left(\frac{2k+1}{2k}\right) = \frac{3}{2} \cdot \frac{5}{4} \cdot \frac{7}{6} \cdots \frac{2n+1}{2n}
   \]

8. **Combine the terms:**
   \[
   \prod_{k=1}^n \left(\frac{2k+1}{2k}\right) = \frac{2n+1}{2} \cdot \frac{2n-1}{2n-2} \cdots \frac{3}{2}
   \]

9. **Final expression:**
   \[
   a_n = \frac{\left(1-\frac1{2^2}\right)\left(1-\frac1{4^2}\right)\left(1-\frac1{6^2}\right)\cdots\left(1-\frac{1}{(2n)^2}\right)}{(2n+1)a_n}
   \]

### Part (c): Find \(\lim_{n\to\infty}a_n\) and justify your answer.

1. **From part (a), we have:**
   \[
   \lim_{n \to \infty} a_n = 0
   \]

2. **Justification:**
   \[
   a_n \approx \frac{e^{-\frac{\gamma}{2}}}{\sqrt{n}}
   \]

3. **As \(n \to \infty\):**
   \[
   \frac{e^{-\frac{\gamma}{2}}}{\sqrt{n}} \to 0
   \]

Thus, the limit is 0.

The final answer is \( \boxed{ 0 } \)