monotonic implication?

Question

let $A_n$ be a monotonic sequence such that $\forall n, A_n \in \mathbb{Z},\,A_n<A_{n+1} $

given the sequence $B_n = (1+{1\over A_n})^{A_n}$

$$\lim \limits_{n \to \infty} B_n = ?$$

now, just from looking at the question, seems like they want me to show $\lim \limits_{n \to \infty} B_n = e$

My answer is:

let $C_n=(1+{1\over n})^n$

from the definition of $A_n$ we can say that

$$\forall n,\, A_n < A_{n+1},\,\, n < n+1$$ $$A_n,n \in \mathbb{Z} \implies \exists N\in \mathbb{Z},\, \forall n>N,\, A_{n+N} = n$$

hence $B_{n+N} = C_n$ and therefore $B_n$ is $C_n$ moved by a finite ($N$) number, so we get
$$\lim \limits_{n \to \infty} B_n = \lim \limits_{n \to \infty} C_n = e$$

Does my proof hold?

Answer 1

Yes, your solution is valid, because any subsequence of a convergent sequence tends to the same value.

Answer 2

It is okay. However，if we drop the condition that $A_n \in Z^+$, it doesn't always hold, for example, let $A_n=1-\frac1n$. Then $\lim B_n=2.$