For a positive integer $n$, $\sigma(n)$ denotes the sum of the positive divisors of $n$. Determine $$\limsup\limits_{n\rightarrow \infty} \frac{\sigma(n^{2023})}{(\sigma(n))^{2023}}$$

[b]Note:[/b] Given a sequence ($a_n$) of real numbers, we say that $\limsup\limits_{n\rightarrow \infty} a_n = +\infty$ if ($a_n$) is not upper bounded, and, otherwise, $\limsup\limits_{n\rightarrow \infty} a_n$ is the smallest constant $C$ such that, for every real $K > C$, there is a positive integer $N$ with $a_n < K$ for every $n > N$.
To determine the value of 
$$ \limsup_{n\rightarrow \infty} \frac{\sigma(n^{2023})}{(\sigma(n))^{2023}}, $$
we start by using the formula for the sum of the divisors function $\sigma(n)$ when $n$ is expressed as a product of prime powers. Let $n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$, then
$$ \sigma(n) = (1 + p_1 + p_1^2 + \cdots + p_1^{\alpha_1})(1 + p_2 + p_2^2 + \cdots + p_2^{\alpha_2}) \cdots (1 + p_k + p_k^2 + \cdots + p_k^{\alpha_k}). $$

1. **Expressing $\sigma(n^{2023})$:**
   For $n^{2023} = (p_1^{\alpha_1})^{2023} (p_2^{\alpha_2})^{2023} \cdots (p_k^{\alpha_k})^{2023}$, we have
   $$ \sigma(n^{2023}) = \sigma(p_1^{2023\alpha_1} p_2^{2023\alpha_2} \cdots p_k^{2023\alpha_k}) = (1 + p_1 + p_1^2 + \cdots + p_1^{2023\alpha_1})(1 + p_2 + p_2^2 + \cdots + p_2^{2023\alpha_2}) \cdots (1 + p_k + p_k^2 + \cdots + p_k^{2023\alpha_k}). $$

2. **Expressing the ratio:**
   We need to evaluate
   $$ \frac{\sigma(n^{2023})}{(\sigma(n))^{2023}} = \frac{(1 + p_1 + p_1^2 + \cdots + p_1^{2023\alpha_1})(1 + p_2 + p_2^2 + \cdots + p_2^{2023\alpha_2}) \cdots (1 + p_k + p_k^2 + \cdots + p_k^{2023\alpha_k})}{[(1 + p_1 + p_1^2 + \cdots + p_1^{\alpha_1})(1 + p_2 + p_2^2 + \cdots + p_2^{\alpha_2}) \cdots (1 + p_k + p_k^2 + \cdots + p_k^{\alpha_k})]^{2023}}. $$

3. **Simplifying the ratio:**
   Consider the ratio for each prime factor $p_i$:
   $$ \frac{1 + p_i + p_i^2 + \cdots + p_i^{2023\alpha_i}}{(1 + p_i + p_i^2 + \cdots + p_i^{\alpha_i})^{2023}}. $$
   Notice that $(1 + p_i + p_i^2 + \cdots + p_i^{\alpha_i})^{2023}$ is a polynomial of degree $2023\alpha_i$ in $p_i$, and $1 + p_i + p_i^2 + \cdots + p_i^{2023\alpha_i}$ is also a polynomial of degree $2023\alpha_i$ in $p_i$.

4. **Asymptotic behavior:**
   For large $n$, the dominant term in both the numerator and the denominator will be the highest power of $p_i$. Thus, asymptotically,
   $$ \frac{1 + p_i + p_i^2 + \cdots + p_i^{2023\alpha_i}}{(1 + p_i + p_i^2 + \cdots + p_i^{\alpha_i})^{2023}} \approx \frac{p_i^{2023\alpha_i}}{(p_i^{\alpha_i})^{2023}} = 1. $$

5. **Conclusion:**
   Since this holds for each prime factor $p_i$, we have
   $$ \frac{\sigma(n^{2023})}{(\sigma(n))^{2023}} \approx 1 $$
   for large $n$. Therefore,
   $$ \limsup_{n\rightarrow \infty} \frac{\sigma(n^{2023})}{(\sigma(n))^{2023}} = 1. $$

The final answer is $\boxed{1}$.