Let $a_n$ be the local maximum of $f_n(x)=\frac{x^ne^{-x+n\pi}}{n!}\ (n=1,\ 2,\ \cdots)$ for $x>0$.

Find $\lim_{n\to\infty} \ln \left(\frac{a_{2n}}{a_n}\right)^{\frac{1}{n}}$.
1. First, we need to find the local maximum of the function \( f_n(x) = \frac{x^n e^{-x + n\pi}}{n!} \) for \( x > 0 \). To do this, we take the derivative of \( f_n(x) \) with respect to \( x \) and set it to zero:
   \[
   f_n'(x) = \frac{d}{dx} \left( \frac{x^n e^{-x + n\pi}}{n!} \right)
   \]
   Using the product rule, we get:
   \[
   f_n'(x) = \frac{e^{-x + n\pi}}{n!} \left( n x^{n-1} - x^n \right) = \frac{e^{-x + n\pi} x^{n-1} (n - x)}{n!}
   \]
   Setting \( f_n'(x) = 0 \), we find:
   \[
   \frac{e^{-x + n\pi} x^{n-1} (n - x)}{n!} = 0
   \]
   This implies \( n - x = 0 \), so \( x = n \).

2. At \( x = n \), \( f_n(x) \) reaches its maximum. Therefore, the local maximum \( a_n \) is:
   \[
   a_n = f_n(n) = \frac{n^n e^{-n + n\pi}}{n!}
   \]

3. We need to find the limit:
   \[
   \lim_{n \to \infty} \ln \left( \frac{a_{2n}}{a_n} \right)^{\frac{1}{n}}
   \]
   First, we express \( a_{2n} \):
   \[
   a_{2n} = f_{2n}(2n) = \frac{(2n)^{2n} e^{-2n + 2n\pi}}{(2n)!}
   \]

4. Now, we compute the ratio \( \frac{a_{2n}}{a_n} \):
   \[
   \frac{a_{2n}}{a_n} = \frac{\frac{(2n)^{2n} e^{-2n + 2n\pi}}{(2n)!}}{\frac{n^n e^{-n + n\pi}}{n!}} = \frac{(2n)^{2n} e^{-2n + 2n\pi} n!}{(2n)! n^n e^{-n + n\pi}}
   \]

5. Taking the natural logarithm and simplifying:
   \[
   \ln \left( \frac{a_{2n}}{a_n} \right) = \ln \left( \frac{(2n)^{2n} e^{-2n + 2n\pi} n!}{(2n)! n^n e^{-n + n\pi}} \right)
   \]
   \[
   = \ln \left( \frac{(2n)^{2n}}{n^n} \right) + \ln \left( \frac{e^{-2n + 2n\pi}}{e^{-n + n\pi}} \right) + \ln \left( \frac{n!}{(2n)!} \right)
   \]
   \[
   = 2n \ln (2n) - n \ln n + (-2n + 2n\pi) - (-n + n\pi) + \ln (n!) - \ln ((2n)!)
   \]
   \[
   = 2n \ln 2 + 2n \ln n - n \ln n - n + n\pi - \ln ((2n)(2n-1)\cdots(n+1))
   \]

6. Using Stirling's approximation \( n! \approx \sqrt{2\pi n} \left( \frac{n}{e} \right)^n \), we get:
   \[
   \ln (n!) \approx n \ln n - n + \frac{1}{2} \ln (2\pi n)
   \]
   \[
   \ln ((2n)!) \approx 2n \ln (2n) - 2n + \frac{1}{2} \ln (4\pi n)
   \]

7. Substituting these approximations:
   \[
   \ln \left( \frac{a_{2n}}{a_n} \right) \approx 2n \ln 2 + 2n \ln n - n \ln n - n + n\pi - (2n \ln (2n) - 2n + \frac{1}{2} \ln (4\pi n))
   \]
   \[
   = 2n \ln 2 + n \ln n - n + n\pi - 2n \ln 2 - 2n \ln n + 2n - \frac{1}{2} \ln (4\pi n)
   \]
   \[
   = -n \ln n + n\pi + n - \frac{1}{2} \ln (4\pi n)
   \]

8. Dividing by \( n \) and taking the limit as \( n \to \infty \):
   \[
   \lim_{n \to \infty} \frac{1}{n} \ln \left( \frac{a_{2n}}{a_n} \right) = \lim_{n \to \infty} \left( -\ln n + \pi + 1 - \frac{1}{2n} \ln (4\pi n) \right)
   \]
   \[
   = \pi - 1
   \]

The final answer is \(\boxed{\pi - 1}\).