This question is almost certainly a duplicate, but I wasn't able to find the original question. If you do find the original question, please feel free to close this as a duplicate.
It is a well-known fact/definition that:
$$ e^a = \lim_{n \to \infty}\left(1 + \frac{a}{n} \right)^n \,,$$
where $a \in \mathbb{R}$ is an arbitrary constant. This is the same as saying that:
$$e^a = \lim_{n \to \infty}\left(1 + \frac{a_n}{n} \right)^n $$ for the constant sequence $a_n \equiv a$. So
Question: For which other sequences (i.e. besides constant sequences) does $$ \lim_{n \to \infty} \left( 1 + \frac{a_n}{n} \right)^n $$ have a limit? And are any of these limits interesting?
Note that a necessary condition is that $a_n = o(n)$. To see this, note that in order for this infinite product to converge we must have
$$ \lim_{n \to \infty} 1 + \frac{a_n}{n } = 1 \quad \iff \quad \lim_{n \to \infty} \frac{a_n}{n} = 0 \quad \iff \quad a_n = o(n).$$
Other than that though I am not sure of much anything.
Note: references suffice for an answer.
Well let's first handle the case when the limit $L$ is positive. Then by taking logs we have $\lim_{n\to\infty} n\log(1+a_n/n)=\log L$ and hence $\lim_{n\to \infty} 1+a_n/n=1$ so that $a_n=o(n) $. Moreover for the limit to be positive we must have $a_n\neq 0$ as $n\to\infty$. Next note that $$\lim_{n\to \infty} n\log\left(1+\frac {a_n} {n} \right) =\lim_{n\to \infty} n\cdot\frac{a_n} {n} \cdot\dfrac{\log\left(1+\dfrac {a_n} {n} \right)} {\dfrac{a_n} {n}} =\lim_{n\to\infty} a_n$$ so that $a_n\to \log L$.
Thus in order that $(1+(a_n)/n)^n\to L>0$ it is necessary as well as sufficient that $a_n\to\log L$.
Note also that $L<0$ is not possible because the sequence is necessarily positive for even $n$.
The only case left is when $L=0$ which is possible (check using $a_n=1-n$). In general if $a_n/n\to m$ and $-2<m<0$ then it is obvious that $L=0$. But I have been unable to find a necessary condition in this case.