Let's break down the problem step by step.

We are given that:

$G_n = \sqrt[n+1]{\prod_{k=1}^{n} C_n^k}$ and $C_n^k = \frac{n!}{k!(n-k)!}$ where $C_n^k$ is the binomial coefficient, and the task is to compute the limit: $\lim_{n \to \infty} \sqrt[n]{G_n}.$

Step 1: Simplify $G_n$

First, let's simplify $G_n$. It is the geometric mean of the binomial coefficients from $C_n^1$ to $C_n^n$. We can write:

$G_n = \sqrt[n+1]{C_n^1 \cdot C_n^2 \cdot \ldots \cdot C_n^n}.$

The binomial coefficient $C_n^k$ is maximized at $k = \frac{n}{2}$. For large $n$, Stirling's approximation for factorials provides a useful way to estimate $C_n^k$. Stirling’s approximation is:

$n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n.$

Using this approximation for the binomial coefficients, we can estimate the behavior of $G_n$ as $n \to \infty$.

Step 2: Estimate $\lim_{n \to \infty} \sqrt[n]{G_n}$

For large $n$, the product of binomial coefficients is dominated by terms around $k \approx \frac{n}{2}$. The geometric mean of these terms will tend towards the central binomial coefficient $C_n^{n/2}$, which behaves asymptotically as:

$C_n^{n/2} \sim \frac{2^n}{\sqrt{\pi n}}.$

Thus, the behavior of $G_n$ for large $n$ is approximately that of $2^n$, and we can write:

$\lim_{n \to \infty} \sqrt[n]{G_n} \sim 2.$

Final Answer:

The limit is: $\boxed{2}.$

Do you want more details or have any questions about this?

Here are 5 related questions to deepen your understanding:

1. How does Stirling’s approximation help in simplifying large factorial expressions?
2. What is the geometric mean and why is it used in this context?
3. How do binomial coefficients behave when $n \to \infty$?
4. Can you explain why the maximum of $C_n^k$ occurs at $k = \frac{n}{2}$?
5. What is the significance of the central binomial coefficient?

Tip: Understanding Stirling’s approximation is crucial for solving problems involving large factorials. It provides an easy way to estimate asymptotic behavior!