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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?

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  • $\begingroup$ If $A_n$ is bounded from above, then the limit of $B_n$ is not $e$ $\endgroup$ – Petite Etincelle Nov 29 '14 at 21:53
  • $\begingroup$ It won't be able to comply with either $A_n \in \mathbb{Z}$ or $A_n < A_{n+1}$ if it is bounded from above. $\endgroup$ – Noobay Nov 29 '14 at 22:00
  • $\begingroup$ Ah you have the conditon $A_n \in \mathbb{Z}$, yeah then it's impossible that $A_n$ is bounded from above $\endgroup$ – Petite Etincelle Nov 29 '14 at 22:01
  • $\begingroup$ something makes me think that saying $A_n$ is a moved series by N needs to be revised to it being a subsequence(how do I prove that?), since given the sequence $A_n = \{1,4500,5130...\}$, $N$ is not a fixed constant and for this case my proof won't hold, or am I wrong? $\endgroup$ – Noobay Nov 30 '14 at 12:54
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Yes, your solution is valid, because any subsequence of a convergent sequence tends to the same value.

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  • $\begingroup$ thanks, I'm new to this and am really insecure about my solutions >,< $\endgroup$ – Noobay Nov 29 '14 at 21:52
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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 $\lim B_n=2.$ 😜

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