# Find a recursive formula for a closed formula recursively at infinity

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))


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 $$a_1=2.$$
• In my limit definition, I let $k \to n$ and $n \to \infty$ so assuming $k = n$ the limit does exist and converges to $e \approx 2.71828$ Feb 12 '20 at 23:13
• @eigenpuss By your definition, $u$ is closed to $+\infty$ for any natural $n$. Feb 12 '20 at 23:18