Convergence of $(1+\frac{1}{n+a_n})^n$, where ${a_n}$ is a real bounded sequence, for $n\to\infty$.

Question

How to approach this proof? My idea for the first step was to use the squeeze theorem:

$\exists_{M \in \mathbb{R}} \forall_{n \in \mathbb{N}} : |a_n| \leq M$, since $a_n$ is bounded, then

$(1+\frac{1}{n+M})^n \leq (1+\frac{1}{n+a_n})^n \leq (1+\frac{1}{n-M})^n$

for all $n \in \mathbb{N}$.

How can I prove that both the left and right expression converge to $e$?

Answer 1

Hint: Suppose that $|a_n|\le A\in\mathbb{Z}$. $$ \underbrace{\lim_{n\to\infty}\left(1+\frac1{n+A}\right)^n}_{\lim\limits_{n\to\infty}\left(1+\frac1{n+A}\right)^{n+A}\lim\limits_{n\to\infty}\left(1+\frac1{n+A}\right)^{-A}} \le\lim_{n\to\infty}\left(1+\frac1{n+a_n}\right)^n \le\underbrace{\lim_{n\to\infty}\left(1+\frac1{n-A}\right)^n}_{\lim\limits_{n\to\infty}\left(1+\frac1{n-A}\right)^{n-A}\lim\limits_{n\to\infty}\left(1+\frac1{n-A}\right)^{A}} $$

Answer 2

Alternative way. Use the fact that $\ln(1+t)=t+o(t)$ as $t\to 0$: $$\begin{align}\left(1+\frac{1}{n+a_n}\right)^n&=\exp\left(n\ln\left(1+\frac{1}{n+a_n}\right)\right)=\exp\left(n\left(\frac{1}{n+a_n}+o(1/n)\right)\right)\\ &=\exp\Big(\frac{1}{1+\underbrace{\frac{a_n}{n}}_{\to 0}}+o(1)\Big)\to \exp(1)=e.\end{align}$$

Answer 3

We know $(1+1/u)^u \to e$ as $u\to \infty.$ Here's a nice thing to know too: If $x_n \to x \in (0,\infty)$ and $y_n \to y \in \mathbb R,$ then $x_n^{y_n} \to x^y.$ So in our problem we can write

$$(1+1/(n+a_n))^n = [(1+1/(n+a_n))^{n+a_n}]^{n/(n+a_n)}.$$

Because $(a_n)$ is bounded, $n+a_n \to \infty$ and $n/(n+a_n) \to 1.$ Hence the expression in brackets $\to e,$  and the exponent $\to 1.$ Apply the above to see the limit is $e^1=e.$