This is false. Consider the sequence

$$a_{j} = \begin{cases}2 && \text{if }j\text{ is odd}\\
\frac{1}{2} && \text{if }j \text{ is even} \end{cases}$$

we have for all $n$

$$1 \leq (a_{1}...a_{n})^{\frac{1}{n}} \leq 2^{\frac{1}{n}}$$

Thus

$$\lim_{n \rightarrow \infty}(a_{1}...a_{n})^{\frac{1}{n}} = 1$$

But

$$\lim_{n \rightarrow \infty}\frac{\sum_{j=1}^{2n}a_{j}}{2n} = \lim_{n \rightarrow \infty}\frac{2n+\frac{1}{2}n}{2n} = \frac{5}{4}.$$

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.

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 $e$.

In particular, the changes from your proof are:

- 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.
- There are no subsequences in my proof, everything follows directly and cleanly from the definitions of limits.
- There is no need for the concept of "shifted" (or "moved", as you call them) sequences.

## Best Answer

Squaring both sides, \begin{align*} a_{n+1}^2 = a_n^2 + 2\frac{a_n}{a_n^3} + \frac{1}{a_n^6} > a_n^2 + 2\frac{1}{a_n^2} \end{align*} and squaring again, \begin{align*} a_{n+1}^4 > a_n^4 + 4 + \frac{4}{a_n^4} > a_n^4 + 4 \end{align*} and so \begin{align*} a_{n+1}^4 = a_1^4 + \sum_{k=1}^{n}(a_{k+1}^4 - a_k^4) > 4n + a_1^4 \rightarrow \infty \end{align*}