In Statistical Inference by George Casella, Lemma 2.3.14 states that:
$\text{Let }a_1,a_2,...\text{be a sequence of numbers converging to }a\text{, that is, }\lim_{n \to \infty}a_n=a\text{. Then}$ $$\lim_{n \to \infty}\left(1+\frac{a_n}{n}\right)^n = e^a$$
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.
The accepted answer should not be regarded as a formal proof because it just postpones the proof of $\lim_{n\to\infty}(1+\frac{a_n}{n})^n=e^a$ to the proof of $\lim((1+\frac{1}{m_n})^{m_n})^{a_n}=(\lim(1+\frac{1}{m_n})^{m_n})^{\lim_{n\to\infty}a_n}$.
First, by the continuity of the natural logarithm, this is equivalent to showing that $\lim_{n\to\infty}n\log(1+\frac{a_n}{n})=a$. Since $\lim_{n\to\infty}a_n=a$, for sufficiently large $n$, it holds that $|a_n - a|\leq 1$.
Then for $n>2(|a_n|+1)$, it holds that $|\frac{a_n}{n}|\leq 2$. Now, when $|x|\leq \frac{1}{2}$, the natural log function is bounded by $x-\frac{x^2}{2}\leq \log(1+x)\leq x$, which gives $a_n - \frac{1}{2}\frac{a_n^2}{n} \leq n\log(1+\frac{a_n}{n})\leq a_n$. As $n$ goes to $\infty$, both the upper bound and the lower bound converge to $a$. Thus $\lim_{n\to\infty}n\log(1+\frac{a_n}{n})=a$
With logarithm using derivative looks elegant - https://math.stackexchange.com/questions/1242720/how-to-prove-the-limit-of-the-exponential-of-a-sequence/1242767#1242767
– nagvalhm Dec 14 '23 at 23:16Define 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 $$\lim_{n\to \infty}(1+\frac1{n})^{n}=e.$$
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 $a_n \to a$, $\dfrac{a_n}{n} \to 0$ and there exists an $N_{0}$ such that $\left|\dfrac{a_n}{n}\right| < \dfrac{1}{2}$ for $n > N_0$. Thus for $n > N_0$, we have by the Taylor series of $\ln(1+x)$ and the triangle inequality that:
$0 \leq\left|\ln\left(1 + \dfrac{a_n}{n}\right) - \dfrac{a_n}{n}\right| \leq \left|\dfrac{a_n}{n}\right|^2 + \left|\dfrac{a_n}{n}\right|^3 + \left|\dfrac{a_n}{n}\right|^4 + ....< \left|\dfrac{a_n}{n}\right|^2\left(1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8}...\right) \leq 2\left|\dfrac{a_n}{n}\right|$. Since $\displaystyle \lim_{n \to \infty} \left|\dfrac{a_n}{n}\right| = 0$, the squeeze theorem implies that $\displaystyle \lim_{n \to \infty} \left(\ln\left(1 + \dfrac{a_n}{n}\right) - \dfrac{a_n}{n}\right) = 0$ which in turn implies that $\displaystyle \lim_{n \to \infty} \dfrac{\ln \left(1 + \dfrac{a_n}{n}\right)}{\dfrac{a_n}{n}} = 1$, and this implies that $ \displaystyle \lim_{n \to \infty} \ln y_n = \displaystyle \lim_{n \to \infty} a_n = a$, and finally that $\displaystyle \lim_{n \to \infty} y_n = e^a$.
By the Mean Value Theorem, for sufficiently large $n$ (such that $1+\frac{a_n}{n} \neq 0$), there is a number $b_n$ between $1$ and $1+\frac{a_n}{n}$, i.e., $$\left\lvert b_n -1 \right\rvert \leq \left\lvert \frac{a_n}{n} \right\rvert = \left\lvert \left(1+\frac{a_n}{n}\right) - 1 \right\rvert,$$ such that $$\ln\left(1+\frac{a_n}{n}\right) - \ln(1) = \frac{1}{b_n} \cdot \frac{a_n}{n}.$$ Then, $$n\ln\left(1+\frac{a_n}{n}\right) \rightarrow \frac{a}{1} = a$$ as $n \rightarrow \infty$. Therefore, by the continuity of $\exp$, $$\left(1+\frac{a_n}{n}\right)^n = e^{n\ln\left(1+\frac{a_n}{n}\right)}\rightarrow e^a$$ as $n \rightarrow \infty$, as desired.
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.
