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The provided image contains three mathematical expressions involving summations and limits. Let's break down each component step-by-step:

### 1. \( F(n) \)
The function \( F(n) \) is given by a double summation:

\[
F(n) = \sum_{m=1}^{n} \left( \binom{n}{m} m! \sum_{k=0}^{m} \frac{(-1)^k}{k!} \right) + 1
\]

- \( \binom{n}{m} \) represents the binomial coefficient, which counts the ways to choose \( m \) elements from a set of \( n \) elements.
- The inner sum involves \( (-1)^k \) divided by \( k! \), which is part of the exponential series expansion for \( e^{-1} \).
- The outer sum runs from \( m = 1 \) to \( n \), multiplying the binomial coefficient \( \binom{n}{m} \) by \( m! \).

### 2. \( G(n) \)
The second expression defines \( G(n) \):

\[
G(n) = \sum_{k=0}^{n} \binom{2n}{2k} \binom{2n - 2k}{n - k} \binom{2k}{k}
\]

- Here, \( \binom{2n}{2k} \), \( \binom{2n - 2k}{n - k} \), and \( \binom{2k}{k} \) are binomial coefficients.
- The sum runs from \( k = 0 \) to \( n \), making this expression another combinatorial identity that is likely related to counting lattice paths or combinatorial structures.

### 3. Limit expression
The third part of the expression is a limit:

\[
\lim_{n \to \infty} \left( \frac{\sqrt{G(n) F(n)}}{n^n} \right)^{\frac{1}{n}} = ?
\]

This is the limit of the expression involving \( G(n) \) and \( F(n) \) as \( n \to \infty \), normalized by \( n^n \). The \( n^n \) term suggests that we are looking for the asymptotic behavior of \( F(n) \) and \( G(n) \) as \( n \) grows large.

### To solve:
- We would need to find the asymptotic behavior of \( F(n) \) and \( G(n) \), potentially using approximations for large \( n \).
- Then, apply these approximations to evaluate the limit.

Would you like a detailed calculation of this limit step-by-step or further explanation on specific components of these sums? 

### Relative Questions:
1. How do we evaluate asymptotic behavior of summations as \( n \to \infty \)?
2. What are common techniques for approximating large factorials and binomial coefficients?
3. How does Stirling's approximation help in analyzing large factorials in summations?
4. Can we simplify combinatorial sums like \( G(n) \) using known identities?
5. What is the significance of the \( n^n \) normalization in the limit?

### Tip:
When dealing with large summations involving factorials and binomial coefficients, asymptotic approximations like Stirling's approximation can simplify the analysis significantly.

Find the limit of the expression involving G(n) and F(n), defined through combinatorial sums, as n approaches infinity.

Explore the asymptotic behavior of the functions F(n) and G(n) defined by combinatorial sums, and find the limit as n approaches infinity. This solution includes binomial coefficients, factorial approximations, and advanced asymptotic techniques, suitable for undergraduate students.

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