For what sequences $a_n$ does the sequence $(1+\alpha a_n)^n$ converge?

Question

We know $ (1+\alpha/n)^n \rightarrow e^{\alpha} $ when $n\rightarrow +\infty$.

Suppose we are given a modified version of the problem: $$ \quad (1+\alpha\cdot a_n)^n \tag{1} $$ The question is to choose $a_n$ in a way that the expression (1) has closed form when $n\rightarrow +\infty$. Can we argue that it is necessary to have $a_n \propto \frac{1}{n}$? Or for what families of $a_n$ in the expression (1) is convergent?

Answer 1

If $a_n=\frac{1}{n^{\beta}}$, the limit of $(1+\alpha\cdot a_n)^n $ will be $1$ as soon as $\beta >1$. If $\beta =1$, the limit is $e^\alpha$.If $\beta <1$, the limit is $e^\alpha$ would be undefined.

Answer 2

First we see that a necessary condition to have the limit $e^\alpha$ is that $\lim\limits_{n\to \infty}a_n=0$. Second using Taylor expansion we get

$$\left(1+\alpha a_n\right)^n=\exp(n\ln(1+\alpha a_n))\sim_\infty\exp(\alpha na_n)\xrightarrow{n\to\infty}e^\alpha\iff \lim_{n\to\infty}na_n=1\iff a_n\sim_\infty\frac1n$$