Calculate and prove the limit $\lim\limits_{n\to\infty }(1+\frac{1}{a_n})^{a_n}$ Given $a_n$ is an increasing monotone sequence of integers

Question

$\lim\limits_{n\to\infty }(1+\frac{1}{a_n})^{a_n}$ given $(a_n)$ is an increasing sequence of integers

NOTE - I noticed these 2 questions here (1) and here (2) but I believe that my question is a bit different because these questions answer only the case of $a_1 \geq 0$

according to the information we can understand that if $a_1 \geq 0$ then it is immediately solved as $\lim\limits_{n\to\infty }(1+\frac{1}{a_n})^{a_n} =e$ since then we can look at $a_n$ as in $a_n=n$ and then the limit would be just an identity $\lim\limits_{n\to\infty }(1+\frac{1}{n})^{n}=e$

otherwise what if $a_1 <0$? we will need to show that there exists an $N \in \Bbb N$ such that $a_N \geq 0$ we can also understand that $a_{n+1}-a_n \geq 1$ then for ever $n$ we get $a_{n+1}-a_n = (a_{n+1}-a_n)+(a_n-a_{n+1})...+(a_2 - a_1) \geq 1+1+1...+1 =n$

I do not know how to continue from here.. I usually have more ideas and stuff I tried on my posts but I really cannot figure out what to do here.

thanks for any help and tips! edit - Thanks to all the comments I tried and cannot figure out on how to actually prove it , I do realize what the limit is worth now and why but I am struggling to prove it as I stated before

EDIT (point of the edit is to solve it organized and in a better way according to information I have and from the comments) - Organizing information

a. $(a_n)$ is an increasing sequence therefore we get $(1)$ $a_{n+1} > a_n$ , $(2)$ thanks to nejimban an increasing sequence of integers must tend to $\infty$ $(3)$ an increasing sequence of integers therefore $a_{n+1}-a_n \geq 1$

b. $\lim\limits_{n\to\infty }(1+\frac{1}{n})^{n}=e$

c. Need to prove that there is an $N \in \Bbb N$ such that for every $n>N$ we get $a_n>0$

Solving - posted as an answer to the question

Answer 1

I think your proof can be written out a little more clearly, and with fewer steps, words and symbols. In particular, I will only use one "lemma" before moving to the main proof.

Lemma: For each $n\in\mathbb N$, $a_n \geq a_0 + n$.

Proof: Induction. For $n=0$, the proof is trivial. For general $n$, assume $a_n\geq a_0 + n$. Then, $a_{n+1} > a_n$ and $a_{n+1}$ is an integer, so $$a_{n+1}\geq a_n + 1 \geq (a_0 + n) + 1 = a_0 + (n+1)$$ which concludes the proof.

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Main proof. Let $$b_n=\left(1+\frac1{a_{n'}}\right)^{a_{n'}}.$$

Let $\epsilon > 0$. Then, from the definition of limits and because we know that $$\lim_{n\to\infty}\left(1+\frac1n\right)^n = e$$

we know that there exists some $M\in\mathbb N$ such that, for all $n\in \mathbb N$ such that $n\geq M$, we have $$\left|e-\left(1+\frac1n\right)^n\right|\leq\epsilon.$$

Let $M' = M - a_0$. Let $n'\in\mathbb N$ be such that $n' > M'$. Then, we have $$a_{n'} \geq a_0 + n' \geq a_0 + M' \geq a_0 + M-a_0 \geq M.$$

Therefore, we know that $a_n' \geq M$, which means that

$$|e-b_{n'}|=\left|e-\left(1+\frac1{a_{n'}}\right)^{a_{n'}}\right|\leq \epsilon.$$

Because $n'$ was arbitrary, we know that the inequality above is true for all $n'$. In other words, this proves the statement:

$$\exists M': \forall n'\in \mathbb N: n'>M'\implies |e-b_{n'}|\leq \epsilon$$

Because the choice of $\epsilon$ was arbitrary, we know that the statement above is true for all $\epsilon > 0$, and this is exactly the definition of the limit of the sequence $b_n$ being $e$.

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In particular, the changes from your proof are:

1. There is no splitting of cases when $a_1$ is positive or negative. This splitting of cases is not needed, because in both cases, we will need $a_n$ to eventually be not only positive, but also larger than some (usually big) number.
2. There are no subsequences in my proof, everything follows directly and cleanly from the definitions of limits.
3. There is no need for the concept of "shifted" (or "moved", as you call them) sequences.

Answer 2

Hopefully this is the right answer after all the amazing help in the comments and hopefully this can help someone if they struggled with this like I did. For now I will accept my own answer as it seems logical to me but if someone finds something wrong with it please let me know

First we organize given information , conclusions and theorems we will use.

1. Given $(a_n)$ is an increasing sequence of integers therefore $a_{n+1}-a_{n}\geq1$
2. $a_{n+1} > a_n$
3. theorem : an increasing sequence which is not bounded must tend to $\infty$
4. $\lim\limits_{n\to\infty }(1+\frac{1}{n})^{n}=e$ is a known limit
5. a subsequence of a convergent sequence also converges to the original series limit
6. sequence $(b_n)$ is called a subsequence of $a_n$ if there is an increasing sequence $(n_k)$ of natural numbers that satisfies $a_{n_k}=b_k$ for every $k \in \Bbb N$

Solution:

We will first prove that $\lim\limits_{n\to\infty }a_n=\infty$ for the case $a_1<0$ (if $a_1 \geq 0$ there isn't much to prove). let $a_1 = -k \in \Bbb Z$.

therefore $a_{n+t} \geq t+a_1=0$ (as podiki said) we are checking $a_1 <0$ so we picked a first element which is an integer so let $a_1 = -t$ . if $a_1 = -t$ then since the series is increasing after $t$ terms we must have a nonnegative number since the sequence goes up by at least one each term then we get $a_{n+t} \geq 0$. now let $N \in \Bbb N$ such that for all $n>N$ we get $a_n >a_N \geq0$ and we found such an $n$ ( $a_{n+t} \geq 0$ ) according to point 2 and point 1 we get that $a_{n+t+1} > a_{n+t}$ therefore according to point 6 this is an increasing sequence of natural numbers. then we can say that $(1+ \frac{1}{a_{n+t}})^{a_{n+t}}$ is a subsequence of $(1+ \frac{1}{n})^n$ and since according to point 4 we know that $\lim\limits_{n\to\infty }(1+\frac{1}{n})^{n}=e$ then $\lim\limits_{n\to\infty }(1+ \frac{1}{a_{n+t}})^{a_{n+t}}=e$ as well, and since $\lim\limits_{n\to\infty }(1+ \frac{1}{a_{n+t}})^{a_{n+t}}=e$ is a moved sequence of $a_n$ (sorry if it is not called "moved sequence" in english hope it is still understood) we get that $\lim\limits_{n\to\infty }(1+ \frac{1}{a_{n}})^{a_{n}}=e$