# Proof for $\lim_{n \to \infty}(1+\frac{a_n}{n})^n = e^a$?

In Statistical Inference by George Casella, Lemma 2.3.14 states that:

$\text{Let }a_1,a_2,...\text{be a sequence of numbers converging to }a\text{, that is, }\lim_{n \to \infty}a_n=a\text{. Then}$ $$\lim_{n \to \infty}(1+\frac{a_n}{n})^n = e^a$$

Can somebody give me some hint for this? My first instinct is that since n is in the denominator, then that limit term would just be 1. However, this is obviously not correct.

• Have you seen the proof that $\displaystyle\lim_{n \to \infty}\left(1+\dfrac{x}{n}\right)^n = e^x$? If not, I recommend reading that one first. – JimmyK4542 Oct 20 '14 at 6:22
• @JimmyK4542, is this what you're referring to? – C.J. Jackson Oct 20 '14 at 6:27

Define the auxiliary variable $m_n=n/a_n$.

Then $$\lim_{n \to \infty}(1+\frac{a_n}{n})^n=\lim_{n\to \infty}(1+\frac1{m_n})^{m_na_n}$$ $$=\lim_{n\to \infty}\left((1+\frac1{m_n})^{m_n}\right)^{a_n}$$ $$=\left(\lim_{n\to \infty}(1+\frac1{m_n})^{m_n}\right)^{\lim_{n\to \infty}a_n}.$$ The exponent tends to $a$. As the sequence of $m_n$ has an accumulation point at $\infty$,the other limit is the same as the well-known $$\lim_{n\to \infty}(1+\frac1{n})^{n}=e.$$

First, try to show that if $\displaystyle\lim_{n \to \infty}a_n = a$, then $\displaystyle\lim_{n \to \infty} n \ln\left(1+\dfrac{a_n}{n}\right) = a$.

You will find the taylor series $\ln(1+x) = x - \dfrac{1}{2}x^2 + \dfrac{1}{3}x^3 - \cdots$ helpful.

After showing the above, exponentiate both sides and you will be done.

For $|x|< 0.5$, we have the estimate $|\log(1+x)-x| \le 2 x^2$.

Hence we have $|n\ln(1+{a_n \over n})- a_n| \le 2 n ( {a_n \over n})^2 = 2 {a_n^2 \over n}$, as long as $|a_n| < {n \over 2}$. Since $a_n \to a$, we see that $|a_n| \le B$ for some $B$, hence if $n > 2B$, then we have $|n\ln(1+{a_n \over n})- a_n| \le 2 {B^2 \over n}$. Consequently, we have $|n\ln(1+{a_n \over n})- a| \le |n\ln(1+{a_n \over n})- a_n| + |a_n -a|$, and so we see that $n\ln(1+{a_n \over n}) \to a$.

Finally, since $(1+{a_n \over n})^n = e^{ n\ln(1+{a_n \over n}) }$ and $\exp$ is continuous, we have $(1+{a_n \over n})^n \to e^a$.

Let $y_n = \left(1 + \dfrac{a_n}{n}\right)^n$, then $\ln y_n = n\ln \left(1 + \dfrac{a_n}{n}\right) = a_n\cdot \dfrac{\ln\left(1+\dfrac{a_n}{n}\right)}{\dfrac{a_n}{n}}$. The main thing is to prove that : $\displaystyle \lim_{n \to \infty} \left(\ln \left(1 + \dfrac{a_n}{n}\right) - \dfrac{a_n}{n}\right) = 0$.

Since $a_n \to a$, $\dfrac{a_n}{n} \to 0$, there exists an $N_{0}$ such that when $n > N_0$, $\left|\dfrac{a_n}{n}\right| < \dfrac{1}{2}$. Thus for $n > N_0$, we have:

$0 \leq\left|\ln\left(1 + \dfrac{a_n}{n}\right) - \dfrac{a_n}{n}\right| \leq \left|\dfrac{a_n}{n}\right|^2 + \left|\dfrac{a_n}{n}\right|^3 + \left|\dfrac{a_n}{n}\right|^4 + ....< \left|\dfrac{a_n}{n}\right|^2\left(1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}...\right) \leq 2\left|\dfrac{a_n}{n}\right|$, and since $\displaystyle \lim_{n \to \infty} 2\left|\dfrac{a_n}{n}\right| = 0$, Squeeze theorem implies $\displaystyle \lim_{n \to \infty} \left(\ln\left(1 + \dfrac{a_n}{n}\right) - \dfrac{a_n}{n}\right) = 0$ which in turn implies $\displaystyle \lim_{n \to \infty} \dfrac{\ln \left(1 + \dfrac{a_n}{n}\right)}{\dfrac{a_n}{n}} = 1$, and this implies that $\displaystyle \lim_{n \to \infty} \ln y_n = \displaystyle \lim_{n \to \infty} a_n = a$, and finally $\displaystyle \lim_{n \to \infty} y_n = e^a$.
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