General formula of $\lim_{n \to \infty} (1+a_n)^{b_n}$ where $b_n \to \infty$

Question

What is the general formula of $$\lim_{n \to \infty} (1+a_n)^{b_n} \quad \text{where} \quad \lim_{n \to \infty} b_n = \infty?$$

For example we have $$\lim_{n\to \infty} \left(1 + \frac1n\right)^n = e$$

Now I want a general formula for $\lim_{n \to \infty} (1+a_n)^{b_n}$. I know if $a_n \to A \neq 0$ then the answer is infinity. But I am not sure about what will happen when $a_n \to 0$. Maybe the answer is $e^c$ where $c = \lim _{n \to \infty} b_n a_n$. But even if this is the answer, I want the proof.

It seems it must be a famous problem but unfortunately, I couldn't find the answer in math.SE. If an answer exists with proof, please post its link in the comments and I will delete my question.

Answer 1

Let $c_n := a_n b_n$ and assume that $0\lt a_n\rightarrow 0$, $b_n\rightarrow \infty$. Then define: $$ A_n:={\left(1+a_n\right)^{b_n}}=\left({\left(1+a_n\right)^{a_n^{-1}}}\right)^{c_n} $$ From this we can clearly observe that the limit of $A_n$ indeed depends on $c_n$. In particular, if $c_n \rightarrow B$, for some $B$, then $A_n\rightarrow e^{B}$. Otherwise, if $c_n$ is divergent, then so is $A_n$.

Answer 2

If the limit does exist, we take logs of both sides, getting $$ \begin{split} \lim_{n \to \infty} (1+a_n)^{b_n} &= \lim_{n \to \infty} \exp\left(b_n \ln (1+a_n)\right) \\ &= \exp \left( \lim_{n \to \infty} b_n \ln (1+a_n)\right) \\ &= \exp \left( \lim_{n \to \infty} \frac{\ln (1+a_n)}{1/b_n}\right) \end{split} $$ So if $a_n \to 0$, we end up with a $0/0$ form, and L'Hospital's rule applies, and the rest depends on the particular dependencies of $a_n$ and $b_n$ on $n$.

UPDATE courtesy of Andrew in comments below

Since $a_n \to 0$, we can use L'Hospital's rule to see that $$ \lim_{n \to \infty} \frac{\ln (1+a_n)}{a_n} = \lim_{u \to 0} \frac{\ln (1+u)}{u} = \lim_{u \to 0} \frac{1}{1+u} = 1. $$ Therefore, $$ \begin{split} \lim_{n \to \infty} \frac{\ln (1+a_n)}{1/b_n} &= \lim_{n \to \infty} \frac{\ln (1+a_n)}{a_n} [a_n b_n] \\ &= \left[\lim_{n \to \infty} \frac{\ln (1+a_n)}{a_n}\right] \left[\lim_{n \to \infty} a_n b_n \right] \\ &= \lim_{n \to \infty} (a_n b_n) \end{split} $$ and therefore, $$ \lim_{n \to \infty} (1+a_n)^{b_n} = \exp \left(\lim_{n \to \infty} \frac{\ln (1+a_n)}{1/b_n} \right) = \exp \left(\lim_{n \to \infty} (a_n b_n)\right) $$

As the example mentioned in the question, if $a_n=1/n, b_n=n$ so $a_n \cdot b_n = 1$, we would get from our result $(1+1/n)^n \to e^1 = e$, as expected.

Answer 3

Let $b_n,a_n$ such that $\lim_n a_nb_n=c$. Moreover assume that $a_n\rightarrow 0$.

Then we have

$$(1+a_n)^{b_n} = (1+a_n)^{\frac{c}{a_n}} \cdot (1+a_n)^{b_n-\frac{c}{a_n}}$$

We deal with each term alone:

First, since $(1+\frac{1}{n})^n\rightarrow e$ we have that $(1+\frac{1}{n})^{cn}\rightarrow e^c$. As $a_n\rightarrow 0$ we can substitute $a_n$ with $\frac{1}{n}$.

It is therefore enough to show that $(1+a_n)^{b_n-\frac{c}{a_n}}\rightarrow 0$.

For every $\varepsilon>0$ there exists $N$ such that for all $n>N$ we have $|a_nb_n-c|<\varepsilon$. This implies that $|b_n-\frac{c}{a_n}|<\frac{\varepsilon}{a_n}$.

Therefore $(1+a_n)^{b_n-\frac{c}{a_n}} < (1+a_n)^{\frac{\varepsilon}{a_n}}$ and the latter convergence to $e^\varepsilon$. As $\varepsilon$ is arbitrary small, $e^\varepsilon$ is arbitrary close to $1$ and this completes the proof.