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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))
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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 $a_1=2.$

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  • $\begingroup$ I already know that, I was trying to define it recursively by formalizing my limit expression. $\endgroup$
    – explogx
    Feb 12 '20 at 23:09
  • $\begingroup$ @eigenpussy It's something another. See please better my post. $\endgroup$ Feb 12 '20 at 23:11
  • $\begingroup$ In my limit definition, I let $k \to n$ and $n \to \infty$ so assuming $k = n$ the limit does exist and converges to $e \approx 2.71828$ $\endgroup$
    – explogx
    Feb 12 '20 at 23:13
  • $\begingroup$ @eigenpuss By your definition, $u$ is closed to $+\infty$ for any natural $n$. $\endgroup$ Feb 12 '20 at 23:18
  • $\begingroup$ Yes, I think you can't take a limit for a recursive formula that involves another limit. That is just not formally correct. I refer about the statement I wrote in my post. $\endgroup$
    – explogx
    Feb 12 '20 at 23:20

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