Find a recursive formula for a closed formula recursively at infinity

Question

I have a recursive sequence defined as such:

$$ \left(u_k \right) = \begin{cases} u_0 = 1 \\ u_k = u_{k-1} + u_{k-1} \cdot \frac{1}{n} \end{cases}\quad \text{with}\ k,n \in \mathbb{N} $$

I want $n$ to be as big as possible (close to infinity), and then $k$ would range from $[0;n]$ to compute the $n$-th term. I wrote:

$$\lim \limits_{\substack{n \to \infty\\k \to n}} u_k = \lim \limits_{\substack{n \to \infty\\k \to n}} \left( u_{k-1} + u_{k-1} \cdot \frac{1}{n} \right)$$

Can I write it this way?

This question is motivated by this limit which correspond to the closed formula of what I am trying to achieve recursively:

$$\lim \limits_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$$

This is similar to this little python script which uses recursion to compute $e$ if you choose $n$ to be large enough:

def euler(n):
    un = 1
    for i in range(1, n + 1):
        un = un + un * 1/n
    return un

print(euler(100000))

Answer 1

It's a geometric progression. $$u_k=u_0\left(1+\frac{1}{n}\right)^k=\left(1+\frac{1}{n}\right)^k.$$ Now, we see that the limit does not exist.

If you want a recursive formula for $a_n=\left(1+\frac{1}{n}\right)^n,$ so we can write something as this: $$a_{n+1}=\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{\left(1+\frac{1}{n}\right)^n}a_n,$$ where $a_1=2.$